Download SCHOOL OF DISTANCE EDUCATION B. Sc. MATHEMATICS MM5B06: ABSTRACT ALGEBRA STUDY NOTES

SCHOOL OF DISTANCE EDUCATION
B. Sc. MATHEMATICS
MM5B06: ABSTRACT ALGEBRA
(Core Course)
FIFTH SEMESTER
STUDY NOTES
Prepared by:
Vinod Kumar P.
Assistant Professor
P. G.Department of Mathematics
T. M. Government College, Tirur
UNIVERSITY OF CALICUT
B.Sc. MATHEMATICS
MM5B06: ABSTRACT ALGEBRA
Study Notes
Prepared by:
Vinod Kumar P.
Assistant Professor
P. G.Department of Mathematics
T. M. Government College, Tirur
Email: [email protected]
Published by:
SCHOOL OF DISTANCE EDUCATION
UNIVERSITY OF CALICUT
June, 2013
Copy Right Reserved
SRINIVASA RAMANUJAN
(December 22, 1887 to April 26, 1920)
“As are the crests on the heads of peacocks,
As are the gems on the hoods of cobras,
So is Mathematics, at the top of all sciences.”
The Yajurveda, circa 600 B.C.
A Note to the Students
———————————————
Mathematics is an abstract branch of human knowledge. Its language
and style are precise and systematic. No one can learn this subject without capturing its language and the way of its thought processes. As Paul Halmos said,
the only way to learn Mathematics is to do Mathematics and you cannot
understand this subject by just reading some books, or by attending some lecture
classes on Mathematics or just by watching someone doing it! So don’t just read
this notes, but try to ask your own questions, find out your own examples, and
thereby try to internalize the concepts. While studying a theorem and its proof,
try to convince yourself where the proof uses the hypothesis and how we arrive at
the final conclusions. Each and every step of the proof requires a thorough analysis, sound reasoning, and explanations. Also, consider the following questions:
Is the entire hypothesis necessary? Is there any other alternative ways to reach
the conclusion? Can we connect the present result with any previous ideas? Is
the converse true?... Such an analytic approach will help you for a better understanding of the concept and to enjoy the pleasure of doing Mathematics. Take
special care to do all the problems listed in these notes, that will give you
much confidence for future studies and to face the exams. Doing problems in
an analytic and systematic way helps to internalize the abstract mathematical
concepts more better. Always remember that success is never an accident,
it is the final out come of purposeful activities and hard work.
Queries and suggestions are most welcome and that can be mailed to: [email protected]
C.U.Campus,
06 - 06- 2013.
Vinod Kumar. P
Contents
Module-I
1
Introduction to Groups
5
7
1.1
Binary Algebraic Structures . . . . . . . . . . . . . . . . . . . . .
1.2
Groups: Definition and Elementary Properties . . . . . . . . . . . 17
1.3
Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Module-II
2
Groups of Permutations
31
2.1
Elementary Properties of Cyclic Groups . . . . . . . . . . . . . . 31
2.2
Groups of Permutations . . . . . . . . . . . . . . . . . . . . . . . 37
2.3
Orbits, Cycles, and the Alternating Groups . . . . . . . . . . . . . 43
Module-III
3
Cosets and the Theorem of Lagrange
50
3.1
Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.2
Theorem of Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . 54
4 Homomorphisms
59
4.1
Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . 59
4.2
Properties of Homomorphisms . . . . . . . . . . . . . . . . . . . . 62
Module-IV
5 Rings, Integral Domains and Fields
69
5.1
Rings and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
5.2
Integral Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
6 Introduction to Vector Spaces
82
6.1
Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . 83
6.2
Linear Dependence and Independence . . . . . . . . . . . . . . . . 88
6.3
Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . 92
Chapter
1
Introduction to Groups
A group is one of the fundamental objects of study in the field of mathematics known as abstract algebra. A group consists of a set of elements and
an operation that takes any two elements of the set and forms another element
of the set in such a way that certain conditions are met. The theory of groups
is the subject of intense study within mathematics, and is used in many scientific fields. The branch of algebra that studies groups is called group theory.
Group theory has extensive applications in mathematics, science, and engineering. Many algebraic structures such as fields and vector spaces may be defined
concisely in terms of groups, and group theory provides an important tool for
studying symmetry, since the symmetries of any object form a group. Groups are
thus essential abstractions in branches of physics involving symmetry principles,
such as relativity, quantum mechanics, and particle physics. Furthermore, their
ability to represent geometric transformations finds applications in chemistry,
computer graphics, and other fields.
As we noted above, group is an algebraic structure consisting of a set together
with an operation that combines any two of its elements to form a third element.
5
Chapter 1.
Introduction to Groups
6
To qualify as a group, the set and the operation must satisfy four conditions called
the group axioms, namely closure, associativity, identity and invertibility. Many
familiar mathematical structures such as number systems obey these axioms:
for example, the integers endowed with the addition operation form a group.
However, the abstract formalization of the group axioms, detached as it is from
the concrete nature of any particular group and its operation, allows entities
with highly diverse mathematical origins in abstract algebra and beyond to be
handled in a flexible way, while retaining their essential structural aspects. The
ubiquity of groups in numerous areas within and outside mathematics makes
them a central organizing principle of contemporary mathematics. The concept
of a group arose from the study of polynomial equations, starting with Évariste
Galois in the 1830’s. After contributions from other fields such as number theory
and geometry, the group notion was generalized and firmly established around
1870.
Many structures investigated in mathematics turn out to be groups. These
include familiar number systems, such as the integers, the rational numbers, the
real numbers, and the complex numbers under addition, as well as the non-zero
rationals, reals, and complex numbers under multiplication. Other important
examples are groups of non-singular matrices (with specified size and type of
entries) under matrix multiplication, and permutation groups, which consist of
invertible functions from a set to itself with composition as group operation.
Group theory allows for the properties of such structures to be investigated in a
general setting.
In what follows, we will discuss in detail the concept of groups with several
illustrating examples. We begin with the definition of binary operations. Recall
that a relation between the sets X and Y is any subset R of X × Y . Also, a
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1.1. Binary Algebraic Structures
7
function or mapping, φ from X to Y is a relation between X and Y such that
each x ∈ X appears as the first member of exactly one ordered pair (x, y) in φ.
If φ : X → Y is a mapping, X is the domain of φ, Y is the codomain of φ,
and the set {φ(x) | x ∈ X}, denoted as φ[X], is the range of φ.
A function φ : X → Y is one to one if φ(x1 ) = φ(x2 ) only when x1 = x2 . The
function φ is onto Y if the range of φ is Y .
Notations.
Z+ , Q+ , and R+ denotes the sets of positive integers, positive rational numbers, and positive real numbers respectively. Also, Q∗ , R∗ , and C∗ denotes the
sets of non zero rational numbers, non zero real numbers, and non zero complex
numbers respectively.
1.1
Binary Algebraic Structures
If m and n are any given integers, we know that the operations addition and
multiplication gives us unique integers m + n and mn respectively. In other
words, addition and multiplication are mappings from Z×Z to Z. Such mappings
are called binary operations. More precisely, we have the following definition.
Definition 1.1.1.
A binary operation ∗ on a set S is a function mapping S × S into S. Thus,
for each (a, b) ∈ S × S, ∗ assigns a unique element of S, denoted as a ∗ b.
We have observed that addition and multiplication are binary operations
on Z. It is clear that these operations defines binary operations on the sets
C, R, R+ , and Z+ . Let R∗ denotes the set of non zero real numbers. Then
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1.1. Binary Algebraic Structures
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addition is not a binary operation on R∗ (Why?), where as multiplication defines a binary operation on R∗ (Why?).
Is division a binary operation on
Z? No, because the quotient of two integers need not be an integer always.
a
is not defined if b = 0. It may be noted that a binary operMoreover,
b
ation on a set S to be defined for every ordered pair (a, b) of elements of S.
For an operation ∗ to be a binary operation on S, we require that (i) exactly
one element is assigned to each possible pair of elements of S, and (ii) for
each ordered pair of elements of S, the element assigned to it is again in S.
Definition 1.1.2.
Let ∗ be a binary operation on S and let H be a subset of S. The H is closed
under ∗ if a ∗ b ∈ H, for all a, b ∈ H. In this case, ∗ induces a binary operation
on H.
For example, addition defines a binary operation on the subset of integers Z
of R. Consider Z with the binary operation subtraction. Clearly, the subset Z+
of positive integers is not closed under subtraction. As another example, consider
Z6 = {0, 1, 2, 3, 4, 5}, with the binary operation addition modulo 6. Clearly the
subsets {0, 3}, and {0, 2, 4} are closed under addition modulo 6. Is there any
other subsets of Z6 which are closed under addition modulo 6?
Problem 1.
Let H = {n2 | n ∈ Z+ } ⊂ Z. Show that H is closed under multiplication. Is
H closed under addition?
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Solution.
Let x, y ∈ H. Then x = n21 , y = n22 for some n1 , n2 ∈ Z+ . Therefore,
xy = n21 .n22 = (n1 .n2 )2 =⇒ xy ∈ H. Hence H is closed under multiplication,but
H is not closed under addition. For instance, 1 = 12 ∈ H, 4 = 22 ∈ H, but
1+4=5∈
/ H.
Definition 1.1.3.
A binary operation on ∗ on a set S is commutative if a ∗ b = b ∗ a for all
a, b ∈ S.
For example, addition and multiplication are commutative binary operations
on Z, whereas subtraction is a binary operation on Z which is not commutative.
On the set Mn (R) of all n × n matrices with entries from R, matrix addition is
a commutative binary operation, but matrix multiplication is not commutative.
Definition 1.1.4.
A binary operation on ∗ on a set S is associative if a ∗ (b ∗ c) = (a ∗ b) ∗ c
for all a, b, c ∈ S.
Clearly, addition and multiplication are associative binary operations in Z,
but subtraction is not associative. Also, matrix addition and matrix multiplication are associative binary operations in Mn (R).
Example 1.
Let S be any non empty set and let P(S) be its power set. It can be easily
checked out that the operations intersection, union, and symmetric difference of
sets are binary operations on P(S). Are they commutative? associative?
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Problem 2.
Let F be the set of all functions from a set S into S. Show that composition
of functions is an associative binary operation on F . Give an example to show
that composition of functions need not be commutative.
Solution.
Let f, g, h ∈ F . Note that f ◦ g is defined as (f ◦ g)(x) = f (g(x)), ∀x ∈ S
=⇒ (f ◦ g) ∈ F . We have, ((f ◦ g) ◦ h)(x) = (f ◦ g)(h(x)) = f (g(h(x))), ∀x ∈ S
and (f ◦ (g ◦ h))(x) = (f (g ◦ h)(x)) = f (g(h(x))), ∀x ∈ S. Thus ((f ◦ g) ◦ h) =
(f ◦ (g ◦ h)) =⇒ Composition of functions is an associative binary operation
in F . Let f, g : R → R be defined as f (x) = sin x and g(x) = 2x. Then
(f ◦g)(x) = f (2x) = Sin(2x) and (g◦f )(x) = g(sin x) = 2 sin x =⇒ f ◦g 6= g◦f .
Thus, composition of functions need not be commutative.
On a finite set, any given binary operation can be represented by means of
a table in which the elements of the set are listed across the top as heads of
columns and at the left side as heads of rows. Note that the elements are listed
on both sides in the same order.
For example, consider the table
*
a
b
c
a
b
c
b
b
a
c
b
c
c
b
a
which defines a binary operation on the set {a, b, c}. The following table
represents the binary operation addition modulo 5 on Z5 .
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+5
0
1 2
3 4
0
0
1 2
3 4
1
1
2 3
4 0
2
2
3 4
0 1
3
3
4 0
1 2
4
4
0 1
2 3
A binary operation defined by table is commutative if and only if
the entries in the table are symmetric with respect to the diagonal
that starts with the upper left corner of the table and ends at the
lower right corner.
The associativity of a binary operation from its table, just by inspection will
not be obvious.
Definition 1.1.5.
A binary algebraic structure hS, ∗i consists of a set S together with a
binary operation ∗ on a set S.
Let hS, ∗i and hS 0 , ∗0 i be two binary algebraic structures. An isomorphism
of S with S 0 is a one to one function φ mapping S onto S 0 such that φ(x ∗ y) =
φ(x) ∗0 φ(y), for all x, y ∈ S.
The property that φ(x ∗ y) = φ(x) ∗0 φ(y), for all x, y ∈ S is called homomorphism property.
If such a mapping φ exists, then S and S 0 are isomorphic binary algebraic
structures, and we denote this as S ' S 0 .
A structural property of a binary algebraic structure is one that must be
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1.1. Binary Algebraic Structures
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shared any isomorphic structure.
For instance, the number of elements in the set S is a structural property of
the binary algebraic structure hS, ∗i, where as the name of the binary operation
(or of the elements ) is not a structural property.
Remark.
If φ : hS, ∗i → hS 0 , ∗0 i is an isomorphism, then φ−1 : S 0 → S exists and φ−1 is
an isomorphism of hS 0 , ∗0 i onto hS, ∗i.
Example 2.
Consider the binary algebraic structures hR, +i with usual addition, and
hR+ , .i with usual multiplication.
Define φ : R → R+ by φ(x) = ex .
If
φ(x) = φ(y), then ex = ey . Taking natural logarithm on both sides we get
x = y, showing that φ is one to one. To see that φ is onto, let r ∈ R+ , then
lnr ∈ R and φ(lnr) = elnr = r. Finally, for x, y ∈ R, we have φ(x + y) = ex+y =
ex .ey = φ(x).φ(y). Thus φ is indeed an isomorphism.
Example 3.
The binary structures hC, .i and hR, .i under usual multiplication are not
isomorphic, since the equation x.x = c has a solution in C for every c ∈ C, but
x.x = −1 has no solution in R. The binary structure hM2 (R), .i of 2 × 2 matrices
of real numbers is not isomorphic to hR, .i, since multiplication of real numbers
is commutative, but matrix multiplication is not commutative.
Problem 3.
Show that the binary structures hQ, +i and hZ, +i under usual addition are
not isomorphic.
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Solution.
Note that the equation x + x = c has a solution in Q for every c ∈ Q, but this
equation may not have solution in Z for every c ∈ Z. For example, x + x = 3
does not have a solution in Z. This shows that these two binary structures are
not isomorphic.
Definition 1.1.6.
Let hS, ∗i be a binary structure. An element e ∈ S is an identity element
for ∗ if e ∗ s = s ∗ e = s, ∀s ∈ S.
Remark.
A binary structure hS, ∗i has at most one identity element. i.e., if the identity
element exists, it is unique.
To see this, let e and ē be identity elements. Then by regarding ē as identity
element, we have e ∗ ē = e. But by regarding e as the identity element, we get
e ∗ ē = ē. This shows that e = ē.
Example 4.
On Z, define the binary operation ∗ as a ∗ b = max{a, b}. Does this operation
has an identity element in Z? If e is the identity element, then a ∗ e = e ∗ a =
a, ∀a ∈ Z =⇒ max{a, e} = a, ∀a ∈ Z =⇒ e ≤ a, ∀a ∈ Z. Since Z has no
smallest element, such an e does not exists.
If we consider the above operation on the set N, 1 is the identity element for ∗.
Problem 4.
On Z × Z, define the binary operation ◦ by (a, b) ◦ (c, d) = (ac, bc + d).
Is ◦ commutative? associative? Find the identity element.
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Solution.
We have (1, 2) ◦ (3, 4) = (3, 10), and (3, 4) ◦ (1, 2) = (3, 6). This shows that ◦
is not commutative. It is an easy exercise to verify that ((a, b) ◦ (c, d)) ◦ (e, f ) =
(a, b) ◦ ((c, d) ◦ (e, f )). Let e = (x, y) be the identity element for ◦. Then
(a, b) ◦ (x, y) = (x, y) ◦ (a, b) = (a, b) =⇒ (ax, bx + y) = (xa, ya + b) =⇒ ax =
xa = a and bx + y = ya + b = b =⇒ (x, y) = (1, 0). Thus (1, 0) is the identity
element for the binary operation ◦.
Theorem 1.1.7.
Suppose hS, ∗i has an identity element e for ∗. If φ : S → S 0 is an isomorphism of hS, ∗i with hS 0 , ∗0 i, then φ(e) is an identity element for the binary
operation ∗0 on S 0 .
Proof.
Let s0 ∈ S 0 . We have to prove that φ(e) ∗ s0 = s0 ∗0 φ(e) = s0 . Since φ is
onto, there exists s ∈ S such that φ(s) = s0 . Since e is the identity element for
∗, we have s ∗ e = e ∗ s = s. Since φ is a function, φ(s ∗ e) = φ(e ∗ s) = φ(e).
Using homomorphism property, we get φ(s) ∗0 φ(e) = φ(e) ∗0 φ(s) = φ(e), which
implies φ(e) ∗0 s0 = s0 ∗0 φ(e) = s0 . Thus, φ(e) is an identity element for the binary
operation ∗0 on S 0 .
Problem 5.
The map φ : Z → Z, defined by φ(n) = n + 1 for n ∈ Z is one to one and onto
Z. Give the definition of a binary operation ∗ on Z such that φ is an isomorphism
mapping
(a) hZ, +i onto hZ, ∗i,
(b) hZ, ∗i onto hZ, +i,
(c) hZ, .i onto hZ, ∗i,
(d) hZ, ∗i onto hZ, .i.
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In each case, find the identity element for ∗ on Z.
Solution.
(a) For φ to be an isomorphism, we must have m ∗ n = φ(m − 1) ∗ φ(n − 1) =
φ((m − 1) + (n − 1)) = φ(m + n − 2) = m + n − 1. The identity element is
φ(0) = 1.
(b) Using the fact that φ−1 must also be an isomorphism, we must have m ∗ n =
φ−1 (m + 1) ∗ φ−1 (n + 1) = φ−1 ((m + 1) + (n + 1)) = φ−1 (m + n + 2) = m + n + 1.
The identity element is φ−1 (0) = −1.
(c) For φ to be an isomorphism, we must have m ∗ n = φ(m − 1) ∗ φ(n − 1) =
φ((m − 1).(n − 1)) = φ(mn − m − n + 1) = mn − m − n + 2. The identity element
is φ(1) = 2.
(d) Using the fact that φ−1 must also be an isomorphism, we must have m ∗ n =
φ−1 (m+1)∗φ−1 (n+1) = φ−1 ((m+1).(n+1)) = φ−1 (mn+m+n+1) = mn+m+n.
The identity element is φ−1 (1) = 0.
Exercises.
1. Determine whether the definition of ∗ give a binary operation on the given
set. If so, check whether ∗ is (i) commutative (ii)associative? Also, examine
∗ for identity element.
(a) On Q, a ∗ b = ab + 1.
(b) On Q, a ∗ b =
ab
.
2
(c) On Z+ , a ∗ b = 2ab .
(d) On Z+ , a ∗ b = ab .
(e) On Z+ , a ∗ b = a − b.
(f) On Z, a ∗ b = max{a, b}.
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(g) On Z, a ∗ b = a + b − ab.
(h) On Z+ , a ∗ b = c, where c is at least 5 more than a + b.
(i) On Z+ , a ∗ b = c, where c is the smallest integer greater than both a
and b.
(j) On Z+ , a ∗ b = c, where c is the greatest integer less than ab.
(k) On Mn R, the set of n × n matrices with real entries under matrix
addition.
(l) On Mn R, the set of n × n matrices with real entries under matrix
multiplication.
2. Find the table representing the binary operation A ∗ B = A ∪ B on the
power set of the set {a, b}.
3. How many binary operations can be defined on a set S with exactly n
elements? How many of them are commutative?
4. Show that every binary operation on a singleton set is both commutative
and associative.
5. Show that every binary operation on a set having just two elements is
associative.
6. Determine whether the given map is an isomorphism from the first binary
structure to the second. Justify your answer. [For problems, (f ) to (j), F
stands for the set of all functions mapping R to R that have derivatives of
all orders.]
(a) hZ, +i with hZ, +i, where φ(n) = −n.
(b) hZ, +i with hZ, +i, where φ(n) = n + 1.
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(c) hQ, .i with hQ, .i, where φ(x) = x2 .
(d) hQ, .i with hQ, .i, where φ(x) = x3 .
(e) hM2 (R), .i with hR, .i, where φ(A) = Determinant of A.
(f) hF, +i with hF, +i, where φ(f ) = f 0 .
(g) hF, +i with hF, +i, where φ(f )(x) =
Rx
(h) hF, +i with hF, +i, where φ(f )(x) =
Rx
d
[
dx 0
0
f (t)dt.
f (t)dt].
(i) hF, .i with hF, .i, where φ(f )(x) = x.f (x).
(j) hF, +i with hR, +i, where φ(f ) = f 0 (0).
7. The map φ : Q → Q, defined by φ(x) = 3x − 1 for x ∈ Q is one to one and
onto Q. Give the definition of a binary operation ∗ on Q such that φ is an
isomorphism mapping
(a) hQ, +i onto hQ, ∗i,
(b)hQ, ∗i onto hQ, +i,
(c) hQ, .i onto hQ, ∗i,
(d)hQ, ∗i onto hQ, .i.
In each case, find the identity element for ∗ on Q.
1.2
Groups: Definition and Elementary Properties
Definition 1.2.1.
A group hG, ∗i is a set G, closed under a binary operation ∗, such that the
following axioms are satisfied.
1. Associativity: (a ∗ b) ∗ c = a ∗ (b ∗ c), ∀a, b, c ∈ G
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1.2. Groups: Definition and Elementary Properties
18
2. Existence of identity element: There is an element e in G such that
e ∗ x = x ∗ e = x, ∀x ∈ G.
3. Existence of inverse element: For each a ∈ G, there is an element
a0 ∈ G such that a ∗ a0 = a0 ∗ a = e.
Here, a0 is called the inverse of a.
Definition 1.2.2.
A group G is abelian if its binary operation is commutative.
A group that is not abelian is called nonabelian.
The word abelian derives from the name of the great Norwegian mathematician Niels Henrik Abel (1802-1829).
Remark.
In a group G , the identity element and inverse of each element are unique.
(Prove this!). For each a ∈ G, the inverse of a is denoted by a−1 .
Example 5.
1. The sets Z, Q, R, and C under addition are abelian groups.
2. The sets Q+ , R+ , Q∗ , R∗ , and C∗ under multiplication are abelian groups.
3. The set Z+ under addition is not a group, since it has no identity element
for + in Z+ .
4. The set Z+ ∪ {0} under addition is not a group, even if it has an identity
element 0, but no inverse for 1.
5. The set Z+ under multiplication is not a group. It has an identity element
1, but no inverse for 2.
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6. The set Mn (R) under matrix addition is an abelian group. The zero matrix
is the identity element.
7. The set Mn (R) under matrix multiplication is not a group, since the zero
matrix has no multiplicative inverse.
Problem 6.
Let hG, ∗i be a group. For a, b ∈ G, prove that (a ∗ b)−1 = b−1 ∗ a−1
Solution.
We have (a∗b)∗(b−1 ∗a−1 ) = a∗(b∗b−1 )∗a−1 = a∗e∗a−1 = a∗a−1 = e. Since the
inverse of any element in a group is unique, this shows that (a ∗ b)−1 = b−1 ∗ a−1 .
Problem 7.
Let ∗ be defined on Q+ by a ∗ b =
ab
. Prove that Q+ is an abelian group
2
under ∗.
Solution.
It is clear that Q+ is closed under ∗. Also, (a ∗ b) ∗ c = a ∗ (b ∗ c) =
abc
,
4
∀a, b, c ∈ Q+ , showing that ∗ is associative.
ab
∗ is commutative, since a ∗ b = b ∗ a = , ∀a, b ∈ Q+ . If e is the identity for
2
ae
∗, then a ∗ e = a =⇒
= a =⇒ e = 2. Finally, computation shows that
2
4
a−1 = , ∀a ∈ Q+ .
a
Theorem 1.2.3.
In a group G, with binary operation ∗ , the left and right cancellation laws
holds. i.e., a ∗ b = a ∗ c =⇒ b = c and b ∗ a = c ∗ a =⇒ b = c, ∀a, b, c ∈ G.
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1.2. Groups: Definition and Elementary Properties
20
Proof.
Suppose a ∗ b = a ∗ c. Multiplying from the left with a−1 on both sides, we
get a−1 ∗ (a ∗ b) = a−1 ∗ (a ∗ c). Using associativity, (a−1 ∗ a) ∗ b = (a−1 ∗ a) ∗ c =⇒
e ∗ b = e ∗ c =⇒ b = c. Similarly,from b ∗ a = c ∗ a, we get b = c.
Theorem 1.2.4.
If G is a group with binary operation ∗ , and if a, b ∈ G the linear equations
a ∗ x = b and y ∗ a = b have unique solutions , x, y in G.
Proof.
Try yourself!
Remark.
A set together with an associative binary operation is called a semigroup.
A monoid is a semigroup that has an identity element for the binary operation.
Note that a group is both a semigroup and a monoid.
Note that there is only one group of a single element, namely {e} with binary
operation e ∗ e = e, up to isomorphism . By looking at the table representations,
we can conclude that there is only one group of two elements ( or three elements)
up to isomorphism . In the next section, we will show that there exists two non
isomorphic group structures on a set of four elements.
Problem 8.
Let S = R \ {−1}. Define ∗ on S by a ∗ b = a + b + ab.
(a) Show that ∗ is a binary operation on S.
(b) Show that hS, ∗i is a group.
(c) Find the solution of the equation 2 ∗ x ∗ 3 = 7 in S.
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1.2. Groups: Definition and Elementary Properties
21
Solution.
(a) We must show that S is closed under ∗, that is, that a + b + ab 6= −1 for
a, b ∈ S. Now a + b + ab = −1 if and only if 0 = ab + a + b + 1 = (a + 1)(b + 1).
This is the case if and only if either a = −1 orb = −1 , which is not the case for
a, b ∈ S.
(b) We have a ∗ (b ∗ c) = a ∗ (b + c + bc) = a + (b + c + bc) + a(b + c + bc) =
a + b + c + ab + ac + bc + abc and (a ∗ b) ∗ c = (a + b + ab) ∗ c = (a + b + ab) + c +
(a + b + ab)c = a + b + c + ab + ac + bc + abc. Thus ∗ is associative. Note that 0
acts as identity element for ∗, since 0 ∗ a = a ∗ 0 = a.
−a
−a
−a
−a
acts as inverse of a, for a ∗
= a+
+a
+1 =
Also, a+1
a+1
a+1
a+1
2
0
a(a + 1) − a − a
=
= 0. Thus hS, ∗i is a group.
a+1
a+1
(c) Because the operation is commutative, 2 ∗ x ∗ 3 = 2 ∗ 3 ∗ x = 11 ∗ x.
−11
Now the inverse of 11 is
, by Part(b). From, 11 ∗ x = 7, we obtain
12
−11
−11
−11
−11 + 84 − 77
−4
−1
x=
∗7=
+7+
7=
=
=
.
12
12
12
12
12
3
Problem 9.
Show that if G is a finite group with identity e and with an even number of
elements, then there is a 6= e in G such that a ∗ a = e.
Solution.
Let S = {x ∈ G | x−1 6= x}. Then S has an even number of elements, because
its elements can be grouped in pairs x, x−1 . Because G has an even number of
elements, the number of elements in G but not in S (the set G − S) must be
even. The set G − S is nonempty because it contains e. Thus there is at least
one element of G − S other than e, that is, at least one element other than e
that is its own inverse.
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1.2. Groups: Definition and Elementary Properties
22
Problem 10.
Let G be a group with a finite number of elements. Show that for any a ∈ G,
there exists an n ∈ Z+ such that an = e.
Solution.
Let G has m elements. Then the elements e, a, a2 , a3 , ..., am are not all different, since G has only m elements. If one of a, a2 , a3 , ..., am is e, then we are
done. If not, then we must have ai = aj where i < j. Repeated left cancellation
of a yields e = aj−i .
Problem 11.
Show that if (a ∗ b)2 = a2 ∗ b2 for all a, b in a group G, then G is abelian.
Solution.
We have (a ∗ b) ∗ (a ∗ b) = (a ∗ a) ∗ (b ∗ b), so a ∗ [b ∗ (a ∗ b)] = a ∗ [a ∗ (b ∗ b)]
and left cancellation yields b ∗ (a ∗ b) = a ∗ (b ∗ b). Then (b ∗ a) ∗ b = (a ∗ b) ∗ b
and right cancellation yields b ∗ a = a ∗ b. Thus G is abelian.
Problem 12.
Let G be a group and let g be one fixed element of G. Show that the map ig
defined by ig (x) = gxg −1 for all x in G, is an isomorphism of G with itself.
Solution.
Let a, b ∈ G. If g ∗ a ∗ g −1 = g ∗ b ∗ g −1 , then a = b by group cancellation, so ig is
a one-to-one map. Because ig (g −1 ∗ a ∗ g) = g ∗ g −1 ∗ a ∗ g ∗ g −1 = a, we see that
ig maps G onto G. We have ig (a ∗ b) = g ∗ a ∗ b ∗ g −1 = g ∗ a ∗ (g −1 ∗ g) ∗ b ∗ g −1 =
(g ∗a∗g −1 )∗(g ∗b∗g −1 ) = ig (a)∗ig (b), so ig satisfies the homomorphism property
also, and is thus an isomorphism.
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1.2. Groups: Definition and Elementary Properties
23
Exercises.
1. Show that the subset S of Mn (R) consisting of all invertible n × n matrices
under multiplication is a group. [This group is called the general linear
group of degree n and is denoted by GL(n, R)]
2. Determine whether the binary operation ∗ gives a group structure on the
given set.
(a) On Z, let ∗ be defined by a ∗ b = ab.
(b) On 2Z = {2n | n ∈ Z},let ∗ be defined by a ∗ b = a + b.
(c) On C, let ∗ be defined by a ∗ b = |ab|.
(d) On R∗ , let ∗ be defined by a ∗ b = ab .
(e) On R+ , let ∗ be defined by a ∗ b =
√
ab.
3. Let nZ = {nm | m ∈ Z} . Show that hnZ, +i is a group and hnZ, +i '
hZ, +i.
4. Let G be a group and let a, b ∈ G. Show that (a ∗ b)−1 = a−1 ∗ b−1 if and
only if a ∗ b = b ∗ a .
5. If G is an abelian group, prove that (a ∗ b)n = an ∗ bn for all integers n.
6. Let G be a group and suppose that a ∗ b ∗ c = e for a, b, c ∈ G. Show that
b ∗ c ∗ a = e also.
7. If ∗ is a binary operation on a set S, an element x ∈ S is an idempotent
for ∗ if x ∗ x = x. Prove that a group has exactly one idempotent.
8. Show that every group G with identity e and such that x ∗ x = e, ∀x ∈ G
is abelian.
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1.3. Subgroups
1.3
24
Subgroups
Let n be a positive integer. If a is an element of a group G, written multiplicatively, we denote the product aaa....a for n factors a by an . We let a0 be the
identity element. Also, a−n denotes the product a−1 a−1 a−1 ...a−1 for n factors.
Definition 1.3.1.
If G is a group, then the order |G| of G is the number of elements in G.
Definition 1.3.2.
If a subset H of a group G is closed under the binary operation and if H with
the induced operation from G is itself a group, then H is a subgroup of G. We
denote this by H ≤ G or G ≥ H. Also, H < G or G > H means that H ≤ G
but H 6= G.
Example 6.
1. If G is any group, then the subgroup consisting of G itself is the improper
subgroup of G. All other subgroups of G are proper subgroups. The
subgroup {e} is the trivial subgroup of G. All other subgroups are nontrivial.
2. hZ, +i < hR, +i, but hQ+ , .i is not a subgroup of hR, +i.
3. The nth roots of unity in C form a subgroup Un of the group C∗ of non
zero complex numbers under multiplication.
4. There are two different group structures of order 4. Consider the group
table of Z4 .
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1.3. Subgroups
25
+4
0
1 2
3
0
0
1 2
3
1
1
2 3
0
2
2
3 0
1
3
3
0 1
2
From the table, it is clear that the only proper subgroup of Z4 is {0, 4}.
Another group structure of order 4 is the group V , the Klein 4-group,
which is described by the following table.
V :
e
a
b
c
e
e
a
b
c
a
a
e
c
b
b
b
c
e
a
c
c
b
a
e
Note that V has three proper nontrivial subgroups, {e, a}, {e, b}, and {e, c}.
Now we present a characterization of subgroups.
Theorem 1.3.3.
A subset H of a group G is a subgroup G if and only if (i). H is closed under
the binary operation of G., (ii). the identity element e of G is in H, (iii). for all
a ∈ H it is true that a−1 ∈ H also.
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1.3. Subgroups
26
Theorem 1.3.4.
Let G be a group and let a ∈ G. Then H = {an | n ∈ Z} is a subgroup of G
and is the smallest subgroup of G that contains a, i.e., every subgroup containing
a contains H.
Problem 13.
Show that a non empty subset H of a group G is a subgroup of G if and only
if ab−1 ∈ H for all a, b ∈ H.
Solution.
Let H be a subgroup of G. Then for a, b ∈ H, we have b−1 ∈ H and
ab−1 ∈ H because H must be closed under the induced operation. Conversely,
suppose that H is nonempty and ab−1 ∈ H for all a, b ∈ H. Let a ∈ H. Then
taking b = a, we see that aa−1 = e is in H. Taking a = e, and b = a, we see that
ea−1 = a−1 ∈ H. Thus H contains the identity element and the inverse of each
element. For closure, note that for a, b ∈ H, we also have a, b−1 ∈ H and thus
a(b−1 )−1 = ab ∈ H.
Problem 14.
Let G be a group and let HG = {x ∈ G | xa = ax, ∀a ∈ G}. Show that HG
is an abelian subgroup of G. (HG is called the center of G.)
Solution.
Clearly HG is closed under the operation and e ∈ HG . From xa = ax, we
obtain xax−1 = a and then ax−1 = x−1 a, showing that x−1 ∈ HG , which is thus
a subgroup. Let a ∈ HG . Then ag = ga for all g ∈ G; in particular,ab = ba for
all b ∈ HG because HG is a subset of G . This shows that HG is abelian.
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1.3. Subgroups
27
Definition 1.3.5.
Let G be a group and let a ∈ G. Then the subgroup H = {an | n ∈ Z} is
called the cyclic subgroup of G generated by a, and is denoted by hai.
If the subgroup hai of G is finite, then order of a is the order of |hai| of this
subgroup. Otherwise, we say that a is of infinite order.
Remark.
If a is of finite order m, then m is the smallest positive integer such that
am = e.
Definition 1.3.6.
An element a of a group G generates G and is a generator for G if hai = G.
A group G is cyclic if there is some element a ∈ G that generates G.
Example 7.
1. The group Z4 is and h1i = h3i = Z4 . Also, h2i = {0, 2} and h0i = {0}.
2. The Klein -4 group V is not cyclic, since hei = {e}, hai = {e, a}, hbi = {e, b}
and hci = {e, c}.
3. The group hZ, +i is cyclic. Both 1 and −1 are generators for this group
and they are the only generators. (Why?)
4. Consider the group hZ, +i. Note that the cyclic subgroup generated by
n ∈ Z consists of all multiples of n. i.e., hni = nZ.
Problem 15.
Show that every cyclic group is abelian.
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1.3. Subgroups
28
Solution.
Let G be cyclic and let a be a generator for G. For x, y ∈ G, there exist
m, n ∈ Z such that x = am and y = an . Then xy = am an = am+n = an+m =
an am = yx, so G is abelian.
Problem 16.
Prove that a cyclic group with only one generator can have at most 2 elements.
Solution.
Let B = {e, a, a2 , a3 , ..., an−1 } be a cyclic group of n elements. Then a−1 =
an−1 also generates G, because (a−1 )i = (ai )−1 = an−i for i = 1, 2, ..., n − 1. Thus
if G has only one generator, we must have n − 1 = 1 and n = 2. Of course,
G = {e} is also cyclic with one generator.
Problem 17.
Show that a group with no proper nontrivial subgroups is cyclic.
Solution.
Let G be a group with no proper nontrivial subgroups. If G = {e}, then G
is of course cyclic. If G 6= {e}, then choose a ∈ G such that a 6= e. We know
that hai is a subgroup of G and hai 6= {e}. Because G has no proper nontrivial
subgroups, we must have hai = G. This shows that G is cyclic.
Problem 18.
Let φ : G → G0 be an isomorphism of a group hG, ∗i with a group hG0 , ∗0 i.
Then show that if G is cyclic, so is G0 .
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1.3. Subgroups
29
Solution.
Let a be a generator of G. We claim φ(a) is a generator of G0 . Let b0 ∈ G0 .
Because φ maps G onto G0 , there exists b ∈ G such that φ(b) = b0 . Because a
generates G, there exists n ∈ Z such that b = an . Because φ is an isomorphism,
b0 = φ(b) = φ(an ) = φ(a)n . Thus G0 is cyclic.
Problem 19.
Let H be a subgroup of a group G. For a, b ∈ G, let a ∼ b if and only if
ab−1 ∈ H. Show that ∼ is an equivalence relation on G.
Solution.
We have to prove that ∼ is reflexive, symmetric and transitive. Let a ∈ G.
Then aa−1 = e and e ∈ H, since H is a subgroup. Thus a ∼ a =⇒ ∼ is reflexive.
Let a, b ∈ G and a ∼ b, so that ab−1 ∈ H. Since H is a subgroup, we have
(ab−1 )−1 = ba−1 ∈ H, so b ∼ a =⇒ ∼ is symmetric. Now, let a, b, c ∈ G and
a ∼ b and b ∼ c. Then ab−1 ∈ H and bc−1 ∈ H so (ab−1 )(bc−1 ) = ac−1 ∈ H,
implies a ∼ c. Thus ∼ is transitive.
Exercises.
1. If H and K are subgroups of a group G, then show that H ∩K is a subgroup
of G. Is H ∪ K a subgroup of G?
2. Determine whether the given set of invertible n × n matrices with real
number entries is a subgroup of GL(n, R)
(a) The n × n matrices with determinant 2.
(b) The n × n matrices with no zeros on the diagonal
(c) The n × n matrices with determinant−1.
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1.3. Subgroups
30
(d) The n × n matrices with determinant −1 or 1.
3. Find the order of the cyclic subgroup of Z8 generated by 2.
4. Let G be a group and let a be one fixed element of G. Show that Ha =
{x ∈ G | xa = ax} is a subgroup of G.
5. Let φ : G → G0 be an isomorphism of a group hG, ∗i with a group hG0 , ∗0 i.
If H is a subgroup of G, then show that φ[H] = {φ(h) | h ∈ H} is a
subgroup of G0 .
6. If G is an abelian group and if H = {a ∈ G | a2 = e}, show that H is a
subgroup of G.
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Chapter
2
Groups of Permutations
Permutations are usually studied as combinatorial objects, we will see in
this chapter that they have a natural group structure, and in fact, there is a
deep connection between finite groups and permutations. Before, moving to
permutation groups, we will discuss some more properties of cyclic groups.
2.1
Elementary Properties of Cyclic Groups
We recall the Division Algorithm for Z. If m is a positive integer and n
is any integer, then there exist unique integers q and r such that n = mq + r,
and 0 ≤ r < m. Here, we regard q as the quotient and r as the non negative
remainder when n is divided by m.
Theorem 2.1.1.
A subgroup of a cyclic group is cyclic.
Proof.
Let G = hai and let H ≤ G. If H = {e}, then H = hei is cyclic. If H 6= hei,
31
2.1. Elementary Properties of Cyclic Groups
32
then an ∈ H for some n ∈ Z+ . Choose m as the smallest integer in Z+ such
that am ∈ H. We claim that H = ham i. Let b ∈ H. Since H ≤ G, we have
b = an for some n ∈ Z+ . By division algorithm, there exists integers q and
r such that n = mq + r for 0 ≤ r < m. Then an = amq+r = (am )q ar =⇒
ar = (am )−q an . Since an ∈ H, am ∈ H, and H is a group, (am )−q ∈ H, so that
(am )−q an ∈ H =⇒ ar ∈ H. Since m was the smallest positive integer such that
am ∈ H and 0 ≤ r < m, we have r = 0. Thus n = qm, so that b = an = (am )q
=⇒ H = ham i. Hence a subgroup of a cyclic group is cyclic.
Corollary 2.1.2.
The subgroups of Z under addition are precisely the groups nZ under addition
for n ∈ Z
Let r and s be two positive integers. Then it is an easy exercise to show
that H = {nr + ms | n, m ∈ Z} is a subgroup of hZ, +i. By above corollary, H
must be cyclic. The positive generator d of this cyclic subgroup is the greatest
common divisor (gcd) of r and s. Since d ∈ H, d = nr + ms for some integers
n and m, so that every integer dividing both r and s, must also divide d. Hence
d is indeed the largest number dividing both r and s.
We now describe all cyclic groups up to isomorphism.
Theorem 2.1.3.
Let G be a cyclic group with generator a. If the order of G is infinite , then
G is isomorphic to hZ, +i. If G has finite order n, then G is isomorphic to
hZn , +n i.
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2.1. Elementary Properties of Cyclic Groups
33
Proof.
Assume that the order of G is infinite. Then, for all positive integers m,
am 6= e. We claim that ah 6= ak , whenever h 6= k. Let h > k. Then if
ah = ak =⇒ ah a−k = e =⇒ ah−k = e, which is a contradiction. Hence, every
element of G can be expressed as am , for a unique m ∈ Z. Define φ : G → Z
as φ(ai ) = i. Clearly, φ is well defined, one to one, and onto. Also, φ(ai aj ) =
φ(ai+j ) = i + j = φ(ai ) + φ(aj ). Thus, φ is an isomorphism.
Now we assume that the order of G is finite, so that am = e for some positive
integer m. Choose the smallest positive integer n such that an = e. If s ∈ Z
and s = nq + r for 0 ≤ r < n, then as = anq+r = (an )q ar = eq ar = ar . Also, as
above if 0 < k < h < n and ah = ak , then ah−k = e and 0 < h − k < n, which is
a contradiction to the choice of n. Thus the elements e, a, a2 , a3 , ..., an−1 are all
distinct and precisely these are the all elements of G. Define the map ψ : G → Zn
as ψ(ai ) = i for i = 0, 1, 2, ..., n − 1.. Then ψ is well defined, one to one and onto.
Since an = e, ai aj = ak , where k = i+n j. Thus ψ(ai aj ) = i+n j = ψ(ai )+n ψ(aj ),
showing that ψ is an isomorphism.
Now we present a basic theorem regarding generators of subgroups for the
finite cyclic groups (Proof is left as an exercise to the student).
Theorem 2.1.4.
Let G be a cyclic group with n elements, and generated by a. Let b ∈ G and let
b = as . Then b generates a cyclic subgroup H of G containing
n
d
elements, where
d is the gcd of n and s. Also, has i = hat i if and only if gcd(s, n) = gcd(t, n).
Corollary 2.1.5.
If a is a generator of a finite cyclic group G of order n, then the other generSchool of Distance Education,University of Calicut
2.1. Elementary Properties of Cyclic Groups
34
ators of G are the elements of the form ar , where r is relatively prime to n. i.e.,
gcd(r, n) = 1.
Remark.
For every n ∈ Z+ , we denote the number of positive integers < n and relatively prime to n by φ(n) and is called the Euler function. Thus, if G is a
cyclic group of order n, then there are φ(n) distinct elements in G, each of which
generates G.
Example 8.
Consider hZ7 , +7 i. Clearly 1 is a generator for this cyclic group. Since all
integers 2, 3, ..., 6 are relatively prime to 7, all of these elements are generators
of hZ7 , +7 i. But for the group hZ8 , +8 i, the only generators are 1, 3, 5, and 7.
Since gcd(2, 8) = 2, 2 generates a subgroup of order 8/2 = 4, namely {0, 2, 4, 6}.
Similarly, 4 generates a subgroup of order 8/4 = 2, namely {0, 4}. Can you
identify the subgroup generated by 6?
Example 9.
Let G be a cyclic group of order 12. Can you find the number of generators
for G?. Here, the problem is that how many integers are there relatively prime
to 12 and less than 12. We know that 1, 5, 7, and 11 are the only integers smaller
than 12 and relatively prime to 12. Hence a cyclic group of order 12 will have 4
generators.
Problem 20.
How many generators are there for a cyclic group of order 60?
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2.1. Elementary Properties of Cyclic Groups
35
Solution.
1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53,and 59 are relatively
prime to 60, so there are 16 generators for a cyclic group of order 60.
Problem 21.
How many elements are in the cyclic subgroup of (a) Z30 generated by 25?
(b) Z42 generated by 30?
Solution.
We make use of theorem 2.1.4. (a) Since gcd (25, 30) = 5 and since
30
= 6,
5
h25i has 6 elements.
(b) Since gcd (30, 42) = 6 and since
42
= 7, h30i has 7 elements.
6
Problem 22.
In the following, give an example of a group with the described property or
explain why no example exists.
(a) A finite group that is not cyclic.
(b) An infinite group that is not cyclic.
(c) A cyclic group having only one generator.
(d) An infinite cyclic group having four generators.
(e) A cyclic group in which every element other than identity is a generator
(f ) An finite cyclic group having four generators.
Solution.
(a) The Klein 4-group
(b) hR, +i
(c) Z2
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2.1. Elementary Properties of Cyclic Groups
36
(d) No such example exists. Every infinite cyclic group is isomorphic to hZ, +i
which has just two generators, 1 and -1.
(e) Zp , with p as a prime.
(f ) Z8 has generators 1, 3, 5, and 7.
Problem 23.
Show that a group that has only a finite number of subgroups must be a finite
group.
Solution.
Note that every group is the union of its cyclic subgroups, because every
element of the group generates a cyclic subgroup that contains the element. Let
G has only a finite number of subgroups, and hence only a finite number of cyclic
subgroups. Now none of these cyclic subgroups can be infinite, for every infinite
cyclic group is isomorphic to Z (by theorem 2.1.3) which has an infinite number of
subgroups, namely Z, 2Z, 3Z, .... Such subgroups of an infinite cyclic subgroup
of G would of course give an infinite number of subgroups of G, contrary to
hypothesis. Thus G has only finite cyclic subgroups, and only a finite number of
those. We see that the set G can be written as a finite union of finite sets, so G
is itself a finite set.
Problem 24.
Show that Zp has no proper non trivial subgroups if p is a prime number.
Solution.
All positive integers less than p are relatively prime to p because p is prime,
and hence they all generate Zp . Thus Zp has no proper cyclic subgroups, and
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2.2.
Groups of Permutations
37
thus no proper subgroups, because as a cyclic group, Zp has only cyclic subgroups
(By theorem 2.1.1, a subgroup of a cyclic group is cyclic ).
Exercises.
1. Let a and b be elements of a group G. Show that if ab has finite order n,
then ba also has order n.
2. Let G be a group and suppose a ∈ G generates a cyclic subgroup of order
2 and is the unique such element. Show that ax = xa for all x ∈ G.
3. Let p and q be distinct prime numbers. Find the number of generators of
the cyclic group Zpq .
4. Let p be a prime number. Find the number of generators of the cyclic
group Zpr , where r is an integer ≥ 1.
5. Let G be a cyclic group of order n and H be a cyclic group of order m such
that gcd (m, n) = 1. Then show that G × H is a cyclic group of order mn.
6. Show that a finite group of order n is cyclic if and only if the group contains
an element of order n.
2.2
Groups of Permutations
We recall that a permutation of a set A is a function φ : A → A that is
both one to one and onto.
Consider the operation function composition ◦ on the collection of all permutations of a set A (we call this operation as permutation multiplication). If
σ and τ are any two permutations of a set A, we denote the composition of σ
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2.2.
Groups of Permutations
38
and τ by στ instead of σ ◦ τ . Note that στ is clearly one to one and onto (Prove
this!). Thus permutation multiplication is a binary operation on the collection
of all permutations of a set A.
Remember that the action of στ on A is in right- to -left order; i.e., first
apply τ , and then σ.
Theorem 2.2.1.
Let A be a nonempty set , and let SA denotes the collection of all permutations
of A. The SA is a group under permutation multiplication.
The proof of the above theorem is left as an exercise.
Definition 2.2.2.
Let A be the finite set {1, 2, 3, ..., n}. The group of all permutations of the
set A is the symmetric group on n letters, and is denoted by Sn .
Note that Sn has n! elements.
Example 10.
Let A = {1, 2, 3}. Then we list below the 3! = 6 elements of the symmetric
group on three letters.

 1
ρ0 = 
1

 1
ρ1 = 
2

 1
ρ2 = 
3

2 3 
,
2 3

2 3 
,
3 1

2 3 
,
1 2

 1
µ1 = 
1

 1
µ2 = 
3

 1
µ3 = 
2

2 3 
,
3 2

2 3 
,
2 1

2 3 
.
1 3
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The multiplication table for S3 is shown in the table given below.
ρ0
ρ1
ρ2
µ1
µ2
µ3
ρ0
ρ0
ρ1
ρ2
µ1
µ2
µ3
ρ1
ρ1
ρ2
ρ0
µ3
µ1
µ2
ρ2
ρ2
ρ0
ρ1
µ2
µ3
µ1
µ1
µ1
µ2
µ3
ρ0
ρ1
ρ2
µ2
µ2
µ3
µ1
ρ2
ρ0
ρ1
µ3
µ3
µ1
µ2
ρ1
ρ2
ρ0
From the table, it is clear that this group is not abelian.
It is interesting to note that there is a natural correspondence between the
elements of S3 and the ways in which two copies of an equilateral triangle with
vertices 1, 2, and 3 can be placed, one covering the other with vertices on top of
vertices. Because of this fact, S3 is also the group D3 of symmetries of an
equilateral triangle.. D3 is also called the third dihedral group.
In this context, the permutations ρi corresponds to rotations and µi corresponds to mirror images in bisectors of angles.
Remark.
Any group of at most 5 elements is abelian.
Lemma 2.2.3.
Let G and G0 be groups and let φ : G → G0 be a one-to-one function such
that φ(xy) = φ(x)φ(y) for all x, y ∈ G. Then the image of G under φ, φ[G] =
{φ(g), g ∈ G}, is a subgroup of G0 and φ provides an isomorphism of G with
φ[G].
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Proof.
Let x0 , y 0 ∈ φ[G]. Then there exists x, y ∈ G such that φ(x) = x0 and
φ(y) = y 0 . By assumption, φ(xy) = φ(x)φ(y) = x0 y 0 =⇒ x0 y 0 ∈ φ[G]. Thus
φ[G] is closed under the operation of G0 .
Let e0 be the identity element in G0 . Then, e0 φ(e) = φ(e) = φ(ee) = φ(e)φ(e) =⇒
e0 = φ(e) (by right cancellation in G0 ) =⇒ e0 ∈ φ[G].
Let x0 ∈ G0 . Choose x ∈ G such that φ(x) = x0 . Note that e0 = φ(e) = φ(xx−1 ) =
φ(x)φ(x−1 ) = x0 φ(x−1 ) =⇒ x0−1 = φ(x)−1 ∈ φ[G]. Thus φ[G] is a subgroup of
G0 .
Also, φ is an isomorphism of G onto φ[G], because φ is a one-to-one map of G
onto φ[G] such that φ(xy) = φ(x)φ(y), ∀x, y ∈ G.
The following theorem due to the British mathematician, Arthur Cayley
(1821 − 1895) illustrates the importance of group of permutations.
Theorem 2.2.4. (Cayley’s Theorem)
Every group is isomorphic to a group of permutations.
Proof.
Let G be a group. We show that G is siomorphic to a subgroup of SG .
By above lemma, we need only to show that there exists a one-to-one function
φ : G → SG such that φ(xy) = φ(x)φ(y), ∀x, y ∈ G.
For x ∈ G, let λx : G → G be defined by λx (g) = xg, ∀g ∈ G. Then λx is
one-to-one, because if λx (a) = λx (b), then xa = xb, so by left cancellation, a = b.
Let c ∈ G. Then x−1 c ∈ G, and λx (x−1 c) = x(x−1 c) = c, showing that λx maps
G onto G. Thus λx is a permutation of G.
Define φ : G → SG as φ(x) = λx , ∀x ∈ G. Suppose that φ(x) = φ(y). Then
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λx = λy as functions mapping G into G. In particular λx (e) = λy (e) =⇒ xe =
ye =⇒ x = y. Thus, φ is one-to-one. It remains only to show that φ(xy) =
φ(x)φ(y), i.e. to show that λxy = λx λy . Let g ∈ G. Then, λxy (g) = (xy)g. Also,
(λx λy )(g) = λx (λy (g)) = λx (yg) = x(yg). Thus by associativity, λxy = λx λy .
This completes the proof.
Problem 25.


 1 2 3 4 5 6 
Let σ = 
. Find (a) |hσi|
3 1 4 5 6 2
(b)σ 100 .
Solution.
(a) Starting with 1 and applying σ repeatedly, we see that σ takes 1 to 3 to 4
to 5 to 6 to 2 to 1, so σ 6 is the smallest possible power of σ that is the identity
permutation. It is easily checked that σ 6 carries 2, 3, 4, 5 and 6 to themselves
also, so σ 6 is indeed the identity and |hσi| = 6.
(b) Since σ 6 is the identity permutation, we have


 1 2 3 4 5 6 
σ 100 = (σ 6 )16 σ 4 = σ 4 = 

6 5 2 1 3 4
Problem 26.
Find the number of elements in the set {σ ∈ S5 |σ(2) = 5}.
Solution.
There are 4 possibilities for σ(1), then 3 possibilities for σ(3), then 2 possibilities for σ(4), and then 1 possibility for σ(5). Thus 4 . 3 . 2 . 1 = 24 possibilities
in all, showing that 24 elements will be in the set {σ ∈ S5 |σ(2) = 5}.
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Problem 27.
Give an example of a nonabelian group such that every proper subgroup is
abelian.
Solution.
Consider the nonabelian group S3 . Its proper subgroups are
{ρ0 , ρ1 , ρ2 }, {ρ0 , µ1 }, {ρ0 , µ2 }, {ρ0 , µ3 },
and {ρ0 }, and they are abelian. Thus, every proper subgroup of S3 is abelian. Problem 28.
Show that Sn is a nonabelian group for n ≥ 3.
Solution.
Let n ≥ 3, and let ρ ∈ Sn be defined by ρ(1) = 2, ρ(2) = 3, ρ(3) = 1, and
ρ(m) = m for 3 < m ≤ n. Let µ ∈ Sn be defined by µ(1) = 1, µ(2) = 3, µ(3) = 2,
and µ(m) = m for 3 < m ≤ n. Then ρµ 6= µρ so Sn is not commutative. (Note
that if n = 3, then ρ = ρ1 and µ = µ1 in S3 .)
Exercises.
1. Draw the multiplication table for the group D4 of the symmetries of a
square. (This group is also called the octic group). Identify the subgroups
of this group. Is D4 abelian?
2. Determine whether the following functions from R to R defines a permutation of R.
(a) f1 (x) = x + 1
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(b) f2 (x) = x2
(c) f3 (x) = −x3
(d) f4 (x) = ex
(e) f5 (x) = x3 − x2 − 2x.
3. Let G be a group. Prove that the permutations ρa : G → G, where
ρa (x) = xa for a ∈ G and x ∈ G do form a group isomorphic to G.
4. Show that H = {σ ∈ Sn |σ(1) = 1} is a subgroup of Sn .
5. Show that a function from a finite set S to itself is one-to-one if and only
if it is onto. Is this true when S is infinite?
2.3
Orbits, Cycles, and the Alternating Groups
Let σ be a permutation of a set A. For a, b ∈ A, we let a ∼ b if and only if
b = σ n (a) for some n ∈ Z. Then ∼ defines an equivalence relation on A (Prove!).
So, corresponding to σ, we obtain a partition of A into the equivalence classes
determined by the above defined equivalence relation. These equivalence classes
are called the orbits of σ.
For instance, the orbits of the identity permutation on A are the singleton
subsets of A, since the identity permutation leaves each element of A fixed.
Example 11.
Find the orbits of


 1 2 3 4 5 6 
(a) σ = 

5 1 3 6 2 4


 1 2 3 4 5 6 7 8 
(b) µ = 

2 3 5 1 4 6 8 7
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(a) To find the orbit containing 1, we apply σ repeatedly, we see that 1 → 5 →
2 → 1, so the orbit containing 1 is {1, 2, 5}. Similarly, 3 → 3, and 4 → 6 → 4,
we see that the other orbits of σ are {3} and {4, 6} .
(b) Proceeding as in (a), we see that the orbits of µ are {1, 2, 3, 4, 5}, {6}, and
{7, 8}.
We now identify a special type of permutations.
Definition 2.3.1.
A permutation σ ∈ Sn is a cycle if it has at most one orbit containing more
than one element.
The length of a cycle is the number of elements in its largest orbit.
Two cycles are disjoint if any integer is moved by at most one of these cycles.
Clearly, identity permutation is a cycle of length 1.
We note that a cycle of length n has order n and the order of a permutation
which is expressed as a product of disjoint cycles is the least common multiple
of the lengths of the cycles.
Example 12.


 1 2 3 4 5 6 7 8 
Consider σ = 
 ∈ S8 . The only orbit of σ that
3 2 6 4 5 8 7 1
contains more than one element is {1, 3, 6, 8}. Hence σ is a cycle of length 4. It
can be represented as (1, 3, 6, 8).


 1 2 3 4 5 6 7 8 
In this notation, (1, 5, 4, 6, 8) represents the cycle 

5 2 3 6 4 8 7 1
in S8 .
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Theorem 2.3.2.
Every permutation σ of a finite set is a product of disjoint cycles.
Proof.
Let B1 , B2 , ..., Br be the orbits of σ, and let µi be the cycle defined by
µi (x) = σ(x) for x ∈ Bi and µi (x) = x, otherwise. Then it is clear that
σ = µ1 .µ2 ...µr . Since the orbits are disjoint, being distinct equivalence classes,
the cycles µ1 , µ2 , ..., µr are also disjoint.
Permutation multiplication is not commutative in general. But, multiplication
of disjoint cycles is commutative. Since the orbits of a permutation are unique,
the representation of a permutation as a product of disjoint cycles, none of
which is the identity permutation is unique up to the order of factors.
Problem 29.
Compute the product of cycles (a) (1, 4, 5)(7, 8)(2, 5, 7)(b) (1, 3, 2, 7)(4, 8, 6)
in S8 .
Solution.
(a) Since the cycles are not disjoint, the order of the product can’t be altered.
We see that 1 → 1 →1 → 4, 2 → 5 → 5 → 1.Proceeding like this, we get
 1 2 3 4 5 6 7 8 
(1, 4, 5)(7, 8)(2, 5, 7) = 

4 1 3 5 8 6 2 7
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(b) Here the cycles are disjoint, so theorder of the product does
not matters.
 1 2 3 4 5 6 7 8 
As in (a)., we get (1, 3, 2, 7)(4, 8, 6) = 

3 7 2 8 5 4 1 6
The following example illustrates that the product of two cycles need not be
a cycle.
Example 13.
 Consider the cycles
 (1, 4, 5, 6) and (2, 1, 5) in S6 . We have (1, 4, 5, 6)(2, 1, 5) =
 1 2 3 4 5 6 

. The orbits of this product are {1, 6}, {3} and {2, 4, 5},
6 4 3 5 2 1
so it is not a cycle.


 1 2 3 4 5 6 
Here, (2, 1, 5)(1, 4, 5, 6) = 
, which is also not a cycle.
4 1 3 2 6 5
A cycle of length 2 is called a transposition. A transposition leaves all
elements but two fixed, and maps each of these onto the other.
Problem 30.
Show that any permutation of a finite set of at least two elements is a product
of transpositions.
Solution.
Direct computations shows that
(a1 , a2 , ..., an ) = (a1 , an )(a1 , an−1 )...(a1 , a3 )(a1 , a2 ).
Thus any cycle is a product of transpositions. By theorem 2.3.2, we know that
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every permutation of a finite set is a product of disjoint cycles. This shows
that any permutation of a finite set of at least two elements is a product of
transpositions.
We note that no permutation in Sn can be expressed both as a product of an even number of transpositions and as a product of an odd
number of transpositions.
We classify a permutation of a finite set as even or odd according to whether
it can be expressed as a product of an even number of transpositions or the
product of an odd number of transpositions, respectively.
The identity permutation ι in Sn , where n ≥ 2 is even, because ι = (1, 2)(1, 2).
Example
14.


 1 2 3 4 5 6 7 8 
(a) 
 = (1, 8)(3, 6, 4)(5, 7) = (1, 8)(3, 4)(3, 6)(5, 7),
8 2 6 3 7 4 5 1
so this
 is an even permutation.
 1 2 3 4 5 6 7 8 
(b) 
 = (1, 3, 4)(2, 6)(5, 8, 7) = (1, 4)(1, 3)(2, 6)(5, 7)(5.8),
3 6 4 1 8 2 5 7
so this is an odd permutation.
Problem 31.
Show that every permutation in Sn , where n ≥ 3 can be written as a product
of at most (n − 1) transpositions.
Solution.
Note that (1, 2)(1, 2) is the identity permutation in Sn , and 2 ≤ n − 1 if
n > 2. Because (1, 2, 3, 4, ..., n) = (1, n)(1, n − 1)...(1, 3)(1, 2), we see that a cycle
of length n can be written as a product of n − 1 transpositions.
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Now a permutation in Sn can be written as a product of disjoint cycles, the
sum of whose lengths is ≤ n. If there are r disjoint cycles involved, we see
the permutation can be written as a product of at most n − r transpositions.
Because r ≥ 1, we can always write the permutation as a product of at most
n − 1 transpositions.
We note that for n ≥ 2, the number of even permutation in Sn is the
same as the number of odd permutation; i.e., Sn has n!/2 even permutations and n!/2 odd permutations.
The following theorem shows that the collection of all even permutations of
{1, 2, 3, ..., n} forms a subgroup of order n!/2 of the symmetric group Sn , and
this group is called the alternating group An on n letters.
Note that the set of all odd permutations is not a subgroup of Sn . (Why?)
Theorem 2.3.3.
If n ≥ 2, then the collection of all even permutations of {1, 2, 3, ..., n} forms
a subgroup of order n!/2 of the symmetric group Sn .
Proof.
Clearly the product of two even permutations is again an even permutation.
Since n ≥ 2, the identity permutation ι in Sn , is even, because ι = (1, 2)(1, 2).
Also, note that if σ is expressed as a product of transpositions, the product of
the same transpositions taken in the opposite order is σ −1 . Thus if σ is even,
so is σ −1 . Thus the collection of all even permutations of {1, 2, 3, ..., n} forms a
subgroup of order n!/2 of the symmetric group Sn .
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Exercises.
1. Find all orbits of the permutation (a) σ : Z → Z, where σ(n) = n + 1. (b)
σ : Z → Z, where σ(n) = n + 2


 1 2 3 4 5 6 7 8 
2. Express the permutation 
 in S8 as a product
3 1 4 7 2 5 8 6
of transpositions.
3. Show that every permutation in Sn , where n ≥ 3 that is not a cycle can
be written as a product of at most (n − 2) transpositions.
4. Findthe order of the permutation

 1 2 3 4 5 6 7 8 9 10 11 12 
σ=
 in S12 .
6 1 9 12 4 11 10 8 3 5 2 7
5. Show that every odd permutation in Sn , where n ≥ 3 can be written as a
product of (2n+3) transpositions and every even permutation as a product
of (2n + 8) transpositions .
6. Show that for every subgroup H of Sn for n ≥ 2, either all the permutations
in H are even or exactly half of them are even.
7. Show that the order of a permutation σ ∈ Sn is the least common multiple
(l.c.m.) of the lengths of its disjoint cycles.
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Chapter
3
Cosets and the Theorem of
Lagrange
Lagrange’s theorem is about finite groups and their subgroups. The theorem
is named after Joseph-Louis Lagrange (1736-1813), an Italian mathematician and
astronomer, who made significant contributions to all fields of analysis, number
theory, and classical and celestial mechanics
Before moving to the Lagrange’s theorem and its consequences, we observe
that every subgroup of a group G induces an important decomposition of G.
3.1
Cosets
Let H be a subgroup of a group G, which may be of finite or infinite order.
We exhibit two partitions of G by defining two equivalence relations ∼L and
∼R on G as follows: a ∼L b if and only if a−1 b ∈ H and a ∼R b if and only
if ab−1 ∈ H. It is easy to see that these relations are reflexive, symmetric and
transitive and hence they are equivalence relations on G. Now, let a ∈ G. Then
50
3.1. Cosets
51
the equivalence class (corresponding to ∼L ) containing a consists of all x ∈ G
such that a−1 x ∈ H. =⇒ a−1 x = h for some h ∈ H. =⇒ x = ah for some
h ∈ H. Thus the equivalence class of a under the equivalence relation ∼L is the
set {ah|h ∈ H} which we denote by aH. Similarly, we find that the equivalence
class of a under the equivalence relation ∼R is the set {ha|h ∈ H} and we denote
this set by Ha.
We call aH, the left coset of H containing a, and Ha, the right coset of
H containing a.
For example, consider the subset 3Z of Z. Here we use additive notation,
so the left coset containing the integer n is n + Z. When n = 0, 0 + 3Z =
3Z = {..., −9, −6, −3, 0, 3, 6, 9, ...}. Similarly, the left coset containing the integer
1 is 1 + 3Z = {..., −8, −5, −2, 1, 4, 7, 10, ...} and the left coset containing 2 is
2 + 3Z = {..., −7, −4, −1, 2, 5, 8, 11, ...}. We note that these three left cosets
constitute a partition of Z. Since the group Z is abelian, the left coset n + 3Z
is the same as the right coset 3Z + n, thus the partition of Z into right cosets is
the same. (Generally, the left and the right coset of a subgroup determined by
the same element need not be equal. )
From above example, we have the following observation.
For a subgroup H of an abelian group G, the partition of G into left cosets of
H is the same as the partition of G into right cosets of H.
Let H be a subgroup of G. Now we show that every left coset and right coset
of H have the same number of elements as H. We show this by exhibiting a one
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52
to one map of H onto a left coset gH of H for a fixed element g of G. Define
φ : H → gH by φ(h) = gh, ∀h ∈ H. Since gH = {gh|h ∈ H}, it is clear that φ
is onto. Now let φ(h1 ) = φ(h2 ) for some h1 , h2 ∈ H. =⇒ gh1 = gh2 . By left
cancellation, we get h1 = h2 . Thus φ is one to one. This shows that every left
coset of H have the same number of elements as H. In a similar way, we can
get a one to one map of H onto the right coset Hg. Thus, every coset(left or
right) of a subgroup H of a group G has the same number of elements
as H.
Problem 32.
Find all cosets of the subgroup 4Z of 2Z.
Solution.
Since 2Z is abelian, the left cosets and the right cosets are the same. The left
coset containing the integer 0 is 0 + 4Z = {..., −8, −4, 0, 4, 8, ...} and left coset
containing the integer 2 is 2 + 4Z = {..., −6, −2, 2, 6, 10, ...}. Since these two left
cosets exhausts 2Z, they form a partition of 2Z.
Problem 33.
Find all cosets of the subgroup h4i of Z12 .
Solution.
The cosets are 0+h4i = h4i = {0, 4, 8}, 1+h4i = {1, 5, 9}, 2+h4i = {2, 6, 10},
and 3 + h4i = {3, 7, 11}.
Problem 34.
Let H be a subgroup of a group G such that g −1 hg ∈ H for all g ∈ G and all
h ∈ H. Show that every left coset gH is the same as the right coset Hg.
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53
Solution.
We show that gH = Hg by showing that each coset is a subset of the other.
Let gh ∈ gH where g ∈ G and h ∈ H. Then gh = ghg −1 g = [(g −1 )−1 hg −1 ]g is
in Hg because (g −1 )−1 hg −1 is in H by hypothesis. Thus gH is a subset of Hg.
Now let hg ∈ Hg where g ∈ G and h ∈ H. Then hg = gg −1 hg = g(g −1 hg) is in
gH because g −1 hg ∈ H by hypothesis. Thus Hg is a subset of gH also, which
shows that gH = Hg.
Problem 35.
Let H be a subgroup of a group G. Show that the number of left cosets of
H is the same as the number of right cosets of H.
Solution. We prove this by exhibiting a one-to-one map between the collection
of left cosets of H and the collection of right cosets of H. For any a ∈ G, we
claim that Ha−1 consists of all inverses of elements in aH . For proving this,
note that since H is a subgroup, we have {h−1 |h ∈ H} = H. Therefore, Ha−1 =
{ha−1 |h ∈ H} = {h−1 a−1 |h ∈ H} = {(ah)−1 |h ∈ H}. This proves that Ha−1
consists of all inverses of elements in aH . Define a map φ from the collection of
left cosets of H into the collection of right cosets of H by φ(aH) = Ha−1 . Then φ
is well defined for if aH = bH, then {(ah)−1 |h ∈ H} = {(bh)−1 |h ∈ H}. Because
Ha−1 may be any right coset of H, the map is onto the collection of right cosets.
Because elements in disjoint sets have disjoint inverses, we see that φ is one to
one. Thus there are the same number of left as right cosets of a subgroup H of
a group G.
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Exercises.
1. Find all cosets of the subgroup 4Z of Z.
2. Find all cosets of the subgroup h2i of Z12 .
3. Let H be the subgroup {ρ0 , µ1 } of S3 . Find the partitions of S3 into left
cosets and right cosets of H.
4. Find all cosets of the subgroup h18i of Z36 .
5. Give an example of a subgroup of a group G of order 6 whose left cosets
give a partition of G into just one cell.
6. Give an example of a subgroup of a group G of order 6 whose left cosets
give a partition of G into 6 cells.
3.2
Theorem of Lagrange
From the examples of groups that we have considered so far, we may
observe that the order of a subgroup H of a finite group G is a divisor of the
order of G. This is known as the theorem of Lagrange, and we have the precise
statement as given below.
Theorem 3.2.1.
Let H be a subgroup of a finite group G. Then the order of H is a divisor of
the order of G.
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Proof.
Let G be a group of order n and the subgroup H have order m. Then every coset
of H also has m elements. Let r be the number of cells in the partition of G into
left cosets of H. Then n = rm, so m is a divisor of n.
Corollary 3.2.2.
Every group of prime order is cyclic.
Proof.
Let G be of prime order p, and let a ∈ G and a 6= e. Then the cyclic subgroup
generated by a, hai must contain at least two elements a and e, so that order of
hai ≥ 2 . But by Lagrange’s theorem, order of H must divide the order of G.
This shows that order of hai must be p, so that G is a cyclic group.
We know that the order of an element is the same as the order of the cyclic
subgroup generated by that element. Thus from Lagrange’s theorem, it follows
that the order of an element of a finite group divides the order of the
group.
Definition 3.2.3.
Let H be a subgroup of a group G. The number of left cosets of H in G is
the index (G : H) of H in G.
The index (G : H) of a subgroup H of G may be finite or infinite. If G is a
finite group, then (G : H) is finite and (G : H) = |G|/|H|, since every coset
of H contains |H| elements. But, there are the same number of left as right
cosets of a subgroup H of a group G (See Problem 35), so the index (G : H)
could be equally well defined as the number of right cosets of H in G.
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56
Example 15.
Consider the group Z24 and its cyclic subgroup h3i generated by 3. Since
h3i = {1, 3, 6, 9, 12, 15, 18, 21} has 8 elements, its index (the number of cosets) is
24/8 = 3.
Problem 36.
Let σ = (1, 2, 5, 4)(2, 3) in S5 . Find the index of hσi in S5 .
Solution.
We have, σ = (1, 2, 5, 4)(2, 3) = (1, 2, 3, 5, 4) generates a cyclic subgroup of
S5 of order 5, so its index (the number of left cosets) is 5!/5 = 4! = 24.
Theorem 3.2.4.
Suppose H and K are subgroups of a group G such that K ≤ H ≤ G, and
suppose that (H : K) and (G : H) are both finite. Then (G : K) is finite, and
(G : K) = (G : H)(H : K).
Proof.
Let {ai H | i = 1, 2, ..., r} be the collection of distinct left cosets of H in G
and {bj K | j = 1, 2, ..., s} be the collection of distinct left cosets of K in H.
Then it suffices to prove {(ai bj )K | i = 1, 2, ..., r, j = 1, 2, ..., s} is the collection
of distinct left cosets of K in G. Let g ∈ G and let g be in the left coset ai H
of H. Then g = ai h for some h ∈ H. Let h be in the left coset bj K of K in H.
Then h = bj k for some k ∈ K, so g = ai bj k and g ∈ ai bj K. This shows that
the collection {(ai bj )K | i = 1, 2, ..., r, j = 1, 2, ..., s} includes all left cosets of
K in G. It remains to show the cosets in the collection are distinct. Suppose
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3.2. Theorem of Lagrange
57
that ai bj K = ap bq K, so that ai bj k1 = ap bq k2 for some k1 , k2 ∈ K. Now bj k1 ∈ H
and bq k2 ∈ H. Thus ai and ap are in the same left coset of H, and therefore
i = p and ai = ap . Using group cancellation, we deduce that bj k1 = bq k2 . But
this means that bj and bq are in the same left coset of K, so j = q. Thus,
{(ai bj )K | i = 1, 2, ..., r, j = 1, 2, ..., s} is the collection of distinct left cosets of
K in G.
By Lagrange’s theorem, if there is a subgroup H of a finite group G, then
the order of H divides the order of G.
In general, the converse of this result is not true. It can be shown
that if G is an abelian group of order n and m divides n, then there is always
a subgroup of order m of G. However, by looking at the subgroups of the non
abelian group A4 (which has order 12) one can see that it has no subgroup of
order 6, even though 6 divides 12.
Problem 37.
Let G be a group of order pq, where p and q are prime numbers. Show that
every proper subgroup of G is cyclic.
Solution.
By Lagrange’s theorem, the possible orders for a proper subgroup are p, q,
and 1. Now p and q are primes and every group of prime order is cyclic, and of
course every group of order 1 is cyclic. Thus every proper subgroup of a group
of order pq must be cyclic.
Problem 38.
Show that a group with at least two elements but with no proper nontrivial
subgroups must be finite and of prime order.
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58
Solution.
Let G be of order ≥ 2 but with no proper nontrivial subgroups. Let a ∈
G, a 6= e. Then hai is a nontrivial subgroup of G, and thus must be G itself.
Because every cyclic group not of prime order has proper subgroups, we see that
G must be finite of prime order.
Problem 39.
Show that if H is a subgroup of index 2 in a finite group G, then every left
coset of H is also a right coset of H.
Solution.
Since H is a subgroup of index 2, the partition of G into left cosets of H must
be H and G − H, the compliment of H in G, because G has finite order and H
must have half as many elements as G. For the same reason, this must be the
partition into right cosets of H. Thus every left coset is also a right coset.
Exercises.
1. Find the index of the subgroup hµ1 i in the group S3 .
2. Let µ = (1, 2, 4, 5)(3, 6) in S6 . Find the index of hµi in S6 .
3. Show that if a group G has finite order n, then an = e, ∀a ∈ G.
4. Find the index of the subgroup h18i of the group Z36 .
5. Show that every left coset of the subgroup Z of the additive group of real
numbers contains exactly one element x such that 0 ≤ x < 1.
6. Show that a finite cyclic group of order n has exactly one subgroup of each
order d dividing n, and that these are all the subgroups it has.
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Chapter
4
Homomorphisms
In this chapter, we will discuss maps from a group G to the group G0 which
preserves the group structure.
4.1
Definition and Examples
Definition 4.1.1.
A map φ of a group G into a group G0 is a homomorphism if the homomorphism property φ(ab) = φ(a)φ(b) holds for all a, b ∈ G.
Remark.
For any groups G and G0 , there is always at least one homomorphism φ : g →
G0 namely the trivial homomorphism defined by φ(g) = e0 for all g ∈ G, where
e0 is the identity element of G0 .
Example 16.
Let φ : G → G0 be a group homomorphism of G onto G0 . Then G0 will
be abelian if G is abelian. To see this, let a0 , b0 ∈ G0 . Since φ is onto, there
59
4.1. Definition and Examples
60
exists a, b ∈ G such that φ(a) = a0 and φ(b) = b0 . Then a0 b0 = φ(a)φ(b) =
φ(ab) = φ(ba) = φ(b)φ(a) = b0 a0 , where the third equality is due to the fact
that G is abelian. This shows that G0 is abelian. Thus, this example illustrates
how one can get information about G0 from a given information about G via a
homomorphism φ : G → G0 .
Example 17.
Let F be the additive group of all functions from R into R and let R be
the additive group of real numbers and c be any fixed real number. Define
φc : F → R by φc (f ) = f (c) for f ∈ F . Since the sum of two functions
f and g is the function f + g whose value at x is f (x) + g(x), we see that
φc (f + g) = (f + g)(c) = f (c) + g(c) = φc (f ) + φc (g). Thus φ is a homomorphism
of F into R, and is called the evaluation homomorphism.
Example 18.
Let GL(n, R) be the multiplicative group of all invertible n × n matrices.
Then φ : GL(n, R) → R∗ defined by φ(A) = detA, the determinant of A, for
all A ∈ GL(n, R) is a homomorphism, since det(AB) = det(A)det(B) and since
det(A) 6= 0 for any invertible n × n matrix A.
Example 19.
Consider the additive group Z of integers. For r ∈ Z, define φr (n) = rn for
all n ∈ Z. Then φr (n + m) = r(n + m) = rn + rm = φr (n) + φr (m), so φr is
a homomorphism of Z into itself. When r = 0, we get φ0 , which is the trivial
homomorphism. Similarly, φ1 is the identity map and φ−1 maps Z onto Z.
Problem 40.
Determine whether the given map φ is a homomorphism.
(a) Let φ : Z → R under addition be given by φ(n) = n.
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61
(b) Let φ : R → Z under addition be given by φ(x) = the greatest integer ≤ x.
(c) Let φ : R∗ → R∗ under multiplication be given by φ(x) = |x|.
(d) Let φ : R → R∗ where R is additive and R∗ is multiplicative, be given by
φ(x) = 2x .
Solution.
(a) It is a homomorphism, because φ(m + n) = m + n = φ(m) + φ(n).
(b) It is not a homomorphism, because φ(2.6 + 1.6) = φ(4.2) = 4 but φ(2.6) +
φ(1.6) = 2 + 1 = 3.
(c) It is a homomorphism, because φ(xy) = |xy| = |x||y| = φ(x)φ(y) for x, y ∈ R∗
(d) It is a homomorphism, because φ(x + y) = 2x+y = 2x 2y = φ(x)φ(y) for
x, y ∈ R∗ .
Problem 41.
Let Mn (R) be the additive group of all n × n matrices with real entries, and
let R be the additive group of real numbers. Determine whether the given map
φ is a homomorphism.
(a) Let φ : Mn (R) → R be given by φ(A) = det(A), the determinant of A ∈
Mn (R).
(b) Let φ : Mn (R) → R be given by φ(A) = tr(A), the trace of A ∈ Mn (R).
(The trace of A, tr(A) is the sum of the elements on the main diagonal of A.)
Solution.


 1 0 
(a) No, it is not a homomorphism. Let n = 2 and A = 
 and B =
0 1




 1 1 
 2 1 

, so that A + B = 
. We see that φ(A + B) = det(A + B) =
1 1
1 2
4 − 1 = 3, but φ(A) + φ(B) = det(A) + det(B) = 1 + 0 = 1.
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4.2. Properties of Homomorphisms
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(b) Yes, it is a homomorphism. Let A = (aij ) and B = (bij ) where the element
with subscript ij is in the ith row and j th column. Then φ(A + B) = tr(A + B) =
Pn
Pn
Pn
i=1 (aii + bii ) =
i=1 aii +
i=1 bii = tr(A) + tr(B) = φ(A) + φ(B).
Exercises.
1. Let Sn be the symmetric group on n letters, and let φ : Sn → Z2 be defined
by φ(σ) = 0 if σ is an even permutation and φ(σ) = 1 if σ is an odd
permutation . Show that φ is a homomorphism.
2. Let F be the additive group of continuous functions from [0, 1] into R and
let R be the additive group of real numbers . Show that the map φ : F → R
R1
defined by φ(f ) = 0 f (x)dx for f ∈ F , is a homomorphism.
3. Show that the map γ : Z → Zn defined by γ(m) = r, where r is the
remainder given by the division algorithm when m is divided by n, is a
homomorphism..
4. Let GL(n, R) be the multiplicative group of invertible n × n matrices
with real entries, and let R be the additive group of real numbers. Let
φ : GL(n, R) → R be given by φ(A) = tr(A), the trace of A. Is φ a
homomorphism? Justify your answer.
4.2
Properties of Homomorphisms
In this section, we will look into some structural features of G and G0 that
are preserved under a homomorphism φ : G → G0 . We begin with the following
definition.
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63
Definition 4.2.1.
Let φ be a mapping of as set X into a set Y , and let A ⊆ X and B ⊆ Y . The
image φ[A] of A in Y under φ is {φ(a) | a ∈ A}. The set φ[X] is the range
of φ. The inverse image φ−1 [B] of B in X is {x ∈ X | φ(x) ∈ B}.
The next theorem shows that a homomorphism φ : G → G0 preserves the
identity element, inverses, and subgroups.
Theorem 4.2.2.
Let φ be a homomorphism of a group G into a group G0 .
(a) If e is the identity element in G, then φ(e) is the identity element e0 in G0 .
(b) If a ∈ G, then φ(a−1 ) = φ(a)−1 .
(c) If H is a subgroup of G, then φ[H] is a subgroup of G0 .
(d) If K 0 is a subgroup of G0 , then φ−1 [K 0 ] is a subgroup of G.
Proof.
Let φ be a homomorphism of a group G into a group G0 .
(a) We have φ(a) = φ(ae) = φ(a)φ(e). Multiplying on the left by φ(a)−1 , we see
that e0 = φ(e). Thus φ(e) must be the identity element e0 in G0 .
(b) We have e0 = φ(e) = φ(aa−1 ) = φ(a)φ(a−1 ). This shows that φ(a−1 ) =
φ(a)−1 .
(c) Let H be a subgroup of G and let φ(a) and φ(b) be any two elements in
φ[H]. Then φ(a)φ(b) = φ(ab), so that φ(ab) ∈ φ[H]. Thus φ[H] is closed under
the operation of G0 . Also, e0 = φ(e) ∈ φ[H] and φ(a)−1 = φ(a−1 ) ∈ φ[H]. Thus
φ[H] is a subgroup of G0 .
(d) Similar to (c), try yourself!
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64
Definition 4.2.3.
Let φ : G → G0 be a homomorphism of groups. The subgroup φ−1 [e0 ] = {x ∈
G | φ(x) = e0 } is the kernel of φ, denoted by Ker(φ).
Theorem 4.2.4.
Let φ : G → G0 be a group homomorphism, and let H = Ker(φ). Let a ∈ G.
Then, the set φ−1 [{φ(a)}] = {x ∈ G | φ(x) = φ(a)} is the left coset aH of H,
and is also the right coset Ha of H. Consequently, the two partitions of G into
left cosets and into right cosets of H are the same.
Proof.
We have to show that {x ∈ G | φ(x) = φ(a)} = aH. Suppose that φ(x) =
φ(a). Then φ(a)−1 φ(x) = e0 , where e0 is the identity element of G0 . But by above
theorem, φ(a)−1 = φ(a−1 ), so that φ(a−1 )φ(x) = e0 . Since φ is a homomorphism,
this implies that φ(a−1 )φ(x) = φ(a−1 x), so we get φ(a−1 x) = e0 . But, this means
that a−1 x is in H = Ker (φ), so a−1 x = h for some h ∈ H =⇒ x = ah ∈ aH.
This shows that {x ∈ G | φ(x) = φ(a)} ⊆ aH.
To prove the reverse inclusion, let y ∈ aH, so that y = ah for some h in H. Then,
φ(y) = φ(ah) = φ(a)φ(h) = φ(a)e0 = φ(a) =⇒ y ∈ {x ∈ G | φ(x) = φ(a)}.
Thus, aH ⊆ {x ∈ G | φ(x) = φ(a)}, so that {x ∈ G | φ(x) = φ(a)} = aH.
In a similar way, we can show that {x ∈ G | φ(x) = φ(a)} = Ha.
Corollary 4.2.5.
A group homomorphism φ : G → G0 is a one- to- one map if and only if Ker
(φ) = {e}.
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4.2. Properties of Homomorphisms
65
Proof.
Assume that Ker (φ) = {e}. Then for every a ∈ G, the elements mapped into
φ(a) are precisely the elements of the left coset a{e} = {a}, which shows that φ
is one-to-one.
Conversely, assume that φ is one-to-one. Since a homomorphism preserves
the identity element, we have φ(e) = e0 . Since φ is one-to-one, we see that e is
the only element mapped into e0 by φ, so Ker(φ) = {e}.
Problem 42.
Find Ker(φ) and φ(25) for the homomorphism φ : Z → Z7 such that φ(1) = 4.
Solution.
Note that Ker(φ) = 7Z, because 4 has order 7 in Z7 . We have φ(25) = φ(21 +
4) = φ(21) +7 φ(4) = 0 +7 φ(4) = φ(1) +7 φ(1) +7 φ(1) +7 φ(1) = 4 +7 4 +7 4 +7 4 =
1 +7 1 = 2.
Problem 43.
Find Ker(φ) and φ(18) for the homomorphism φ : Z → Z10 with φ(1) = 6.
Solution.
Note that Ker(φ) = 5Z, because 6 has order 5 in Z10 . We have φ(18) =
φ(15 + 3) = φ(15) +10 φ(3) = 0 +10 φ(3) = φ(1) +10 φ(1) +10 φ(1) = 6 +10 6 +10 6 =
2 +10 6 = 8.
Problem 44.
How many homomorphisms are there of Z onto Z?
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66
Solution.
Because the homomorphism φ must be onto Z, φ(1) must be a generator of
Z. Thus there are only two such homomorphisms φ, one where φ(1) = 1 so
φ(n) = n for all n ∈ Z, and one where φ(1) = −1 so φ(n) = −n for all n ∈ Z. Problem 45.
How many homomorphisms are there of Z into Z?
Solution.
There are an infinite number of homomorphisms from Z into Z. For any
nonzero n ∈ Z, we know that hni is isomorphic to Z, and that φ : Z → Z given
by φ(m) = mn is an isomorphism, and hence a homomorphism. Of course φ
defined by φ(m) = 0 for all m ∈ Z is also a homomorphism.
Definition 4.2.6.
A subgroup H of a group G is normal if its left and right cosets coincide,
that is, if gH = Hg for all g ∈ G.
Remark.
Clearly, all subgroups of abelian groups are normal.
The following corollary is immediate from the theorem 4.2.4.
Corollary 4.2.7.
If φ : G → G0 is a group homomorphism, then Ker(φ) is a normal subgroup
of G.
Problem 46.
Show that any group homomorphism φ : G → G0 where |G| is a prime must
either be the trivial homomorphism or a one-to-one map.
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67
Solution.
By theorem 4.2.2, Ker(φ)= φ−1 [{e0 }] is a subgroup of G. By the Lagrange’s
theorem, either Ker(φ)= {e} or Ker(φ)= G because |G| is a prime number. If
Ker(φ)= {e}, then φ is one-to-one by corollary 4.2.5. If Ker(φ)= G, then φ is
the trivial homomorphism, mapping everything into the identity element.
Problem 47.
Let G be a group. Let h, k ∈ G and let φ : Z × Z → G be defined by
φ(m, n) = hm k n . Show that φ is a homomorphism if and only if hk = kh.
Solution.
We have φ(1, 0) = h1 k 0 = h and φ(0, 1) = h0 k 1 = k. Assume that φ is a
homomorphism. Using addition notation in Z × Z as usual, we have φ(1, 1) =
φ((1, 0)+(0, 1)) = φ(1, 0) φ(0, 1) = hk, φ(1, 1) = φ((0, 1)+(1, 0)) = φ(0, 1) φ(1, 0) =
kh. Thus if φ is a homomorphism, we must have hk = kh.
Conversely, assume that hk = kh. Then for any (i, j) and (m, n) in Z ×
Z , we have φ((i, j) + (m, n)) = φ(i + m, j + n) = hi+m k j+n = hi hm k j k n =
hi k j hm k n = φ(i, j)φ(m, n), where the first equality in the second line follows
from the commutativity of h and k. Thus φ is a homomorphism if and only if
hk = kh
Exercises.
1. How many homomorphisms are there of Z into Z2 ?
2. Let G be a group, and let g ∈ G. Let φg : G → G0 be defined by φg (x) = gx
for x ∈ G. For which g ∈ G is φg a homomorphism?
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68
3. Let G be a group, and let g ∈ G. Let φg : G → G0 be defined by φg (x) =
gxg −1 for x ∈ G. For which g ∈ G is φg a homomorphism?
4. Let φ : G → H be a group homomorphism. Show that φ[G] is abelian if
and only if for all x, y ∈ G, we have xyx−1 y −1 ∈ Ker(φ).
5. Show that if G, G0 and G00 are groups and if φ : G → G0 and γ : G0 → G00
are homomorphisms, then the composite map γφ : G → G00 is a homomorphism.
6. Show that a homomorphism of a cyclic group is completely determined by
its value on a generator of the group.
7. Let G be any group and let a be any element of G. Let φ : Z → G be
defined by φ(n) = an . show that φ is a homomorphism. Describe the image
and the possibilities for the kernel of φ.
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Chapter
5
Rings, Integral Domains and
Fields
A ring is an algebraic structure consisting of a set together with two binary operations usually called addition and multiplication, where the set is an
abelian group under addition and a semi group under multiplication such that
multiplication distributes over addition. One of the most common examples of
a ring is the set of integers endowed with its natural operations of addition and
multiplication. The branch of mathematics that studies rings is known as ring
theory. Ring theorists study properties common to both familiar mathematical
structures such as integers and polynomials, and to the many less well-known
mathematical structures that also satisfy the axioms of ring theory. The ubiquity
of rings makes them a central organizing principle of contemporary mathematics.
Ring theory may be used to understand fundamental physical laws, such as those
underlying special relativity and symmetry phenomena in molecular chemistry.
The study of rings originated from the theory of polynomial rings and the
theory of algebraic integers. In the 1880’s Richard Dedekind introduced the con69
5.1. Rings and Fields
70
cept of a ring, and the term ring was coined by David Hilbert in 1892. After
contributions from other fields, mainly number theory, the ring notion was generalized and firmly established during the 1920’s by Emmy Noether and Wolfgang
Krull. Modern ring theory-a very active mathematical discipline-studies rings
in their own right. To explore rings, mathematicians have devised various notions to break rings into smaller, better-understandable pieces, such as ideals,
quotient rings and simple rings. In addition to these abstract properties, ring
theorists also make various distinctions between the theory of commutative rings
and non commutative rings-the former belonging to algebraic number theory and
algebraic geometry. A particularly rich theory has been developed for a certain
special class of commutative rings, known as fields, which lies within the realm
of field theory. Likewise, the corresponding theory for noncommutative rings,
that of noncommutative division rings, constitutes an active research interest for
noncommutative ring theorists. Since the discovery of a mysterious connection
between noncommutative ring theory and geometry during the 1980’s by Alain
Connes, noncommutative geometry has become a particularly active discipline
in ring theory.
5.1
Rings and Fields
Definition 5.1.1.
A ring hR, +, .i is a set R together with two binary operations + and .,
called addition and multiplication, defined on R such that the following axioms
are satisfied:
(i) hR, +i is an abelian group.
(ii) Multiplication is associative.
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5.1. Rings and Fields
71
(iii) For all a, b, c ∈ R, the left distributive law, a.(b + c) = a.b + a.c and the
right distributive law (a + b).c = a.c + b.c hold.
Notations.
We denote the multiplication in a ring by juxtaposition. i.e., we use ab to
denote a.b
Example 20.
Let R be any ring, and let Mn (R) be the set of all n × n matrices with entries
from R. The addition and multiplication in R allows us to define matrix addition
and multiplication in the usual way. Then it can be shown that Mn (R) is a ring
with these operations. In particular, we can have the rings Mn (Z), Mn (Q), Mn (R)
and Mn (C). Note that multiplication is not a commutative operation in any of
these rings for n ≥ 2.
Example 21.
Let F be the set of all functions from a set R into R. Then F is a ring under
the usual function addition and point wise multiplication defined by (f + g)(x) =
f (x) + g(x) and (f g)(x) = f (x)g(x) for x ∈ R and for all f, g ∈ F .
The following theorem shows that our usual rule of signs are valid in any ring
and its proof is straight forward and we leave it as an exercise to the student.
Theorem 5.1.2.
If R is a ring with additive identity 0, then for any a, b ∈ R we have
1. 0a = a0 = 0
2. a(−b) = (−a)b = −(ab)
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72
3. (-a)(-b)=ab.
Definition 5.1.3.
A map φ from a ring R to a ring R0 is a homomorphism if (i) φ(a + b) =
φ(a) + φ(b) (ii) φ(ab) = φ(a)φ(b) for all a, b ∈ R.
A homomorphism φ : R → R0 from a ring R to a ring R0 is a isomorphism
if it is both one to one and onto .
The rings R and R0 are then isomorphic.
Example 22.
Consider the rings hZ, +, .i and h2Z, +, .i. We know that as abelian groups,
hZ, +i and h2Z, +i are isomorphic under the map φ : Z → 2Z defined by φ(x) =
2x, ∀x ∈ Z. But φ is not a ring isomorphism, for φ(xy) = 2xy, while φ(x)φ(y) =
2x2y = 4xy.
Definition 5.1.4.
A ring in which the multiplication is commutative is a commutative ring.
A ring with a multiplicative identity element is a ring with unity; the multiplicative identity element 1 is called unity.
A multiplicative inverse of an element a in a ring R with unity 1 6= 0 is
an element a−1 ∈ R such that aa−1 = a−1 a = 1.
Let R be a ring with unity 1 6= 0. An element u in R is a unit of R if it has
a multiplicative inverse in R. If every nonzero element of R is a unit, then R is
a divison ring (or skew field).
A field is a commutative division ring. A noncommutative division ring is
called a strictly skew field.
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5.1. Rings and Fields
73
Example 23.
Note that hZ, +, .i is a commutative ring with unity. The only units in it are
−1 and 1. Thus Z is not a field. However, hQ, +, .i, hR, +, .i and hC, +, .i are
fields.
Example 24.
Consider nZ,where n ∈ Z+ with usual addition and multiplication. Then it
is a commutative ring, but without unity unless n = 1, and is not a field.
Example 25.
Z × Z with addition and multiplication by components is a commutative ring
with unit (1, 1), but is not a field because (2, 0) has no multiplicative inverse.
The only units in Z × Z are (1, 1), (1, −1), (−1, 1), and (−1, −1).
Example 26.
Consider Z × Q × Z with addition and multiplication by components . This
is a commutative ring with unity, but is not a field. The units of this ring are
(1, q, 1), (−1, q, 1), (1, q, −1) and (−1, q, −1) for any nonzero q ∈ Q.
Problem 48.
Describe all ring homomorphisms of Z into Z.
Solution.
Let φ : Z → Z be a ring homomorphism. Because 12 = 1, we see that φ(1)
must be an integer whose square is itself, namely either 0 or 1. If φ(1) = 1,
then φ(n) = φ(n1) = n, so φ is the identity map of Z onto itself which is a
homomorphism. If φ(1) = 0, then φ(n) = φ(n1) = 0, so φ maps everything onto
0, which also yields a homomorphism.
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74
We list below some important points to be remembered about rings and fields.
• A multiplicative inverse of an element a in a ring R with unity is unique, if
it exists.
• The set of nonzero elements of a field is an abelian group under multiplication.
• A subring of a ring is a subset of the ring that is a ring under induced
operations from the whole ring.
• A subfield of a field is a subset of the field that is a field under induced
operations from the whole field.
• Unity is the multiplicative identity element, while a unit is an element having
a multiplicative inverse.
• Unity is a unit, but not every unit is a unity.
• The units in Zn are precisely those m ∈ Zn such that gcd(m, n) = 1.
Problem 49.
Consider the map det of Mn (R) into R where det(A) is the determinant of
the matrix A for A ∈ Mn (R). Is det a ring homomorphism? Justify your answer.
Solution.
Because det(A + B) need not equal det(A) + det(B), we see that det is not
a ring homomorphism. For example, det(In + In ) = 2n but det(In ) + det(In ) =
1 + 1 = 2.
Problem 50.
If U is the collection of all units in a ring hR, +, .i with unity, show that hU, .i
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is a group.
Solution.
Let u, v ∈ U . Then there exists s, t ∈ R such that us = su = 1 and
vt = tv = 1. These equations show that s and t are also units in U . Then
(ts)(uv) = t(su)v = t1v = tv = 1 and (uv)(ts) = u(vt)s = u1s = 1, so uv is
again a unit =⇒ U is closed under multiplication. Of course multiplication in U
is associative because multiplication in R is associative. The equation (1)(1) = 1
shows that 1 is a unit. Since any unit u ∈ U has a multiplicative inverse s in U ,
we see that U is a group under multiplication.
Exercises.
1. Show that hnZ, +, .i is a ring.
2. Show that Zn is a ring under the operations of addition modulo n and
multiplication modulo n.
3. Describe all units in the ring (a) Q (b) Z5 .
4. Describe all ring homomorphisms of Z into Z × Z.
5. Describe all ring homomorphisms of Z × Z into Z.
6. Let F be the ring of all functions from R into R. For each a ∈ R, define φa :
F → R by φa (f ) = f (a), ∀f ∈ F . Show that φa is a ring homomorphism.
(This map is called the evaluation homomorphism) .
7. Show that an intersection of subrings of a ring R is again a subring of R.
8. Show that an intersection of subfields of a field F is again a subfield of F .
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9. Let R be a ring, and let a be a fixed element of R. Let Ia = {x ∈ R | ax =
0}. Show that Ia is a subring of R.
5.2
Integral Domains
We know that the product of any two nonzero integers is always nonzero.
But this may not be the case of arbitrary rings. For example, consider the ring
hZ6 , +, .i. Note that product of two nonzero elements may be zero in this ring,
for instance, 2.3 = 0 in Z6 . We call these type of elements as 0 divisors.
Definition 5.2.1.
If a and b are two nonzero elements of a ring R such that ab = 0, then a and
b are divisors of 0(or 0 divisors)
The following theorem illustrates the importance of the concept of 0 divisors
in rings.
Theorem 5.2.2.
The cancellation laws hold in a ring R if and only if R has no divisors of 0.
Proof.
Assume that R is a ring in which the cancellation laws hold, and let ab = 0
for some a, b ∈ R. If a 6= 0, we have ab = a0 =⇒ b = 0, by left cancellation
law. Similarly, if b 6= 0, ab = 0b =⇒ a = 0, by right cancellation law. Thus R
has no divisors of 0, if the cancellation laws hold in R.
Conversely, suppose that R has no divisors of 0, and suppose that ab = ac, with
a 6= 0. Since a 6= 0 and since R has no divisors of 0, we have 0 = ab − ac = a(b −
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c) =⇒ b − c = 0 =⇒ b = c. In a similar way, ba = ca with a 6= 0 =⇒ b = c.
Thus, if R has no divisors of 0, the cancellation laws hold in R.
Definition 5.2.3.
An integral domain D is a commutative ring with unity 1 6= 0 and containing no divisors of 0.
Example 27.
The rings Z and Zp for any prime p are integral domains, but Zn is not an
integral domain if n is not a prime. The direct product of two integral domains D1
and D2 is not an integral domain, because the product of two nonzero elements
may be zero, for instance (d, 0)(0, d0 ) = (0, 0).
We now show that a field has no divisors of 0.
Theorem 5.2.4.
Every field F is an integral domain.
Proof.
Let a, b ∈ F with a 6= 0. Assume that ab = 0. Since a 6= 0, the multiplicative
inverse a−1 of a exists, multiplying the above equation on both sides by a−1 ,
we get a−1 (ab) = a−1 0 = 0. This implies, 0 = a−1 (ab) = (a−1 a)b = eb = b,
which shows that F has no divisors of 0. Since F is a field, in particular F is a
commutative ring with unity, and we showed that F has no divisors of 0. Hence
F is an integral domain.
We know that Z is an integral domain, but not a field. We next prove that
finite integral domains are fields.
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Theorem 5.2.5.
Every finite integral domain is a field.
Proof.
Let D be a finite integral domain, and let 0, 1, a1 , a2 , ..., an be all the distinct
elements of D. To show D is a field, we need to prove that if a ∈ D with
a 6= 0, then the multiplicative inverse a−1 of a exists in D. Now, consider
a1, aa1 , aa2 , ..., aan . All these are distinct elements of D, for if aai = aaj =⇒
a(ai − aj ) = 0 =⇒ ai = aj which is a contradiction to our assumption . Also,
since D has no divisors of 0, none of these elements is 0. Thus the elements
a1, aa1 , aa2 , ..., aan are the elements 1, a1 , a2 , ..., an in some order, so that either
a1 = 1 or aai = 1 for some i. Thus a has a multiplicative inverse.
Since Zp is an integral domain, if p is a prime, from the above theorem, it
follows that Zp is a field if p is a prime.
Definition 5.2.6.
If for a ring R a positive integer n exists such that a + a + ... + a = n.a = 0
for all a ∈ R, then the least such positive integer is the characteristic of the
ring R.
If no such positive integer exists, then R is of characteristic 0.
For example, the ring Zn is of characteristic n, while Z, Q, R and C all have
characteristic 0.
Theorem 5.2.7.
Let R be a ring with unity. If n.1 6= 0 for all n ∈ Z+ , then R has characteristic
0. If n.1 = 0 for some n ∈ Z+ , then the smallest such positive integer is the
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characteristic of R.
Proof.
If n.1 6= 0 for all n ∈ Z+ , then surely we cannot have n.a = 0, ∀a ∈ R for
some n ∈ Z+ . Therefore, R has characteristic 0.
If there is an n ∈ Z+ such that n.1 = 0, then n.a = a + a + ... + a = a(1 + 1 +
... + 1) = a(n.1) = a.0 = 0. Thus, the smallest n ∈ Z+ such that n.1 = 0 is the
characteristic of R.
Example 28.
The ring Z3 × Z3 has characteristic 3 and Z3 × Z4 has characteristic 12. Can
you find the characteristic of Z6 × Z15 ?
Problem 51.
Find all solutions in Z6 of (a) x2 + 2x + 4 = 0, and (b) x2 + 2x + 2 = 0.
Solution.
(a) Trying all possibilities −2, −1, 0, 1, 2, and 3, we find that x = 2 is the only
solution in Z6 .
(b) Trying all possibilities -2, -1, 0, 1, 2, and 3, we find that the given equation
has no solutions in Z6 .
Problem 52.
Let R be a commutative ring with unity of characteristic 3. Compute and
simplify (a + b)9 for a, b ∈ R .
Solution.
Since R has characteristic 3, 3.a = 0 for all a ∈ R, we get
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(a + b)9 = [(a + b)3 ]3 = [a3 + 3.a2 b + 3.ab2 + b3 ]3 = (a3 + b3 )3 = a9 + 3.a6 b3 +
3.a3 b6 + b9 = a9 + b9 .
Problem 53.
Show that the characteristic of an integral domain D must be either 0 or a
prime p.
Solution.
Suppose that the characteristic of D is mn for m > 1 and n > 1.
Then we have (m.1)(n.1) = (mn).1 = 0. Since an integral domain have no
divisors of 0, we must have either m.1 = 0 or n.1 = 0. But if m.1 = 0, then
Theorem 5.2.7 shows that the characteristic of D is at most m. If n.1 = 0, then
characteristic of D is at most n. Thus the characteristic of D cant be a composite
positive integer, so it must either be 0 or a prime p.
Problem 54.
An element a of a ring R is idempotent if a2 = a. Show that a division ring
contains exactly two idempotents.
Solution.
If a2 = a, then a2 − a = a(a − 1) = 0. If a 6= 0, then a−1 exists in R and we
have a − 1 = (a−1 a)(a − 1) = a−1 [a(a − 1)] = a−1 0 = 0, so a − 1 = 0 =⇒ a = 1.
Thus 0 and 1 are the only two idempotent elements in a division ring.
Exercises.
1. A subdomain of an integral domain D is a subset of D that is an integral
domain under induced operations from D. Show that the intersection of
subdomains of an integral domain D is again a subdomain of D.
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2. Find all solutions of the equation x3 − 2x2 − 3x = 0 in Z12 .
3. Solve the equation 3x = 2 in Z7 .
4. Find the characteristic of Z3 × Z.
5. Let R be a commutative ring with unity of characteristic 3. Compute and
simplify (a + b)6 for a, b ∈ R .
6. Show that a finite ring R with unity 1 6= 0 and no divisors of 0, is a division
ring.
7. Show that the characteristic of a subdomain of an integral domain D is
equal to the characteristic of D.
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Chapter
6
Introduction to Vector Spaces
Linear Algebra is that branch of Mathematics which treats the common
properties of algebraic systems that consists of a set, together with a notion of
linear combination of elements in the set.
In this chapter, we study an important algebraic system namely, vector
spaces. Vector spaces are a central theme in modern mathematics and has
extensive applications in the natural sciences and the social sciences. The ideas
behind the abstract notion of a vector space occurred in many concrete examples
during the nineteenth century and earlier. It was Descartes and Fermat who first
discussed the vector spaces R2 and R3 in much the way that are presented today.
In the study of vector spaces, we generalize the geometric concept of a vector
as a line segment of given length and direction advantageously in an abstract
way. The modern definition of a vector space seems to be due to the Italian
mathematician Giuseppe Peano (1858-1932 ). Theory of vector spaces is a fundamental tool in pure and applied Mathematics and is becoming increasingly
important in the Physical, Biological and Social Sciences.
82
6.1. Definition and Examples
6.1
83
Definition and Examples
Definition 6.1.1.
Let a non empty set V of elements be given and a rule of addition of any
two elements of V , denoted by + and another rule of multiplication of elements
of V by real numbers (or complex numbers) also be defined, then the set V is
called a vector space and elements of V are called vectors of V , provided the
following conditions are satisfied.
1. For any u, v ∈ V , addition + is defined and u + v ∈ V . i.e., V is closed under
addition. Further,
(a) Addition is commutative, u + v = v + u for all u, v ∈ V ;
(b) Addition is associative, u + (v + w) = (u + v) + w for all u, v, w ∈ V ;
(c) There is a unique element 0 ∈ V , called the zero vector, such that u + 0 = u
for all u ∈ V (Existence of additive identity);
(d) For each vector u ∈ V , there is a unique vector −u ∈ V , called the additive
inverse of u such that u + (−u) = 0;
2. Multiplication of vectors by scalars (real or complex numbers), satisfies αu ∈
V for each u ∈ V and each scalar α. i.e., V is closed under scalar multiplication.
Further,
(e) If α is a scalar, then α(u + v) = αu + αv, for all u, v ∈ V
(f) If α, β are two scalars, then (α + β)u = αu + βu for all u ∈ V ;
(g) If α, β are two scalars, then α(β)u = (αβ)u = β(αu) for all u ∈ V ;
(h) For all elements u of V , we have 1.u = u. Here, 1 is called the unit scalar.
Remark.
Note that the conditions on the operation addition, in the above definition of
a vector space, says simply that V is an abelian group with respect to addition. If
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u, v are elements of V , then u+(−v) is usually written as u−v. Note that for any
vector v ∈ V , we have 0.v = 0, here the 0 on the left side is the scalar zero, and the
zero on the right side is the zero vector. ( 0.v = (0+0).v = 0.v+0.v =⇒ 0.v = 0).
Similarly, α0 = 0, for any scalar α.
Definition 6.1.2.
A nonempty subset S of a vector space V is called a subspace of V , if S
itself is a vector space with the operations of addition and scalar multiplication
on V .
To check whether S is a subspace or not, we need only to check that S is
closed under addition and scalar multiplication and that 0 ∈ S.
Clearly, if V is any vector space, then V itself is a subspace of V . Also, the
subset consisting of the zero vector alone is a subspace of V , called the zero
subspace of V , and it is the smallest subspace of any vector space V .
The following theorem gives a characterization of subspaces of a vector space.
Theorem 6.1.3.
A non empty subset S of a vector space V is a subspace of V if and only if
(a) If u, v ∈ S, then u + v ∈ S for all u, v ∈ S.
(b) If u ∈ S, then αu ∈ S for each u ∈ S and every scalar α.
Proof.
Assume that (a) and (b) are true. Then commutativity and associativity of
addition follows because elements of S are in particular elements of the vector
space V . Choosing α = −1, from (b), we get −1u ∈ S, for all u ∈ S. Also, by
taking α = 0 in (b), we see that 0u = 0 ∈ S. Therefore, S is a vector space,
hence a subspace of V .
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Conversely, if S is a subspace of V , then S itself is a vector space, and in
particular, S is closed under vector addition and scalar multiplication. Hence S
satisfies the conditions (a) and (b).
The above characterization of subspaces can be modified in the following way.
A non empty subset S of a vector space V is a subspace of V if and only if for
each pair of vectors u, v ∈ S and for each scalar α, the vector αu + v is again
in S.
Now we look at some examples of vector spaces.
Example 29.
Let V be the set of all n-tuples, u = (x1 , x2 , ..., xn ) of real numbers. For
any u = (x1 , x2 , ..., xn ), v = (y1 , y2 , ..., yn ) in V and for any real number α, we
define addition and scalar multiplication as follows: u + v = (x1 + y1 , x2 +
y2 , ..., xn + yn ), αu = (αx1 , αx2 , ..., αxn ). It is easy to see that these definitions of
vector addition and scalar multiplication makes V a vector space. This vector
space is denoted by Rn . Now let W be the subset of V consisting of all
u = (x1 , x2 , ..., xn ) with x1 = 0. Then W is a subspace of V (Prove this!).
Example 30.
Let V be the set of all functions from the set R into R. Define sum of two
elements f, g of V as the function f + g , defined by (f + g)(x) = f (x) + g(x).
The product of the scalar α and the function f is defined as the function αf ,
where (αf )(x) = αf (x). It is an easy exercise to verify these operations make V
a vector space. Let W be the subset of V consisting of all continuous functions
on R and U be the subset of V consisting of all differentiable functions on R .
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Then W and U are subspaces of V . This is because if f and g are continuous
(resp. differentiable) functions on R and α is a real number, then f + g and αf
are continuous( resp. differentiable) functions on R. Also, the zero function is
both continuous and differentiable. In fact, U is a subspace of W , because every
differentiable function is continuous. The space P of polynomial functions on R
is a subspace of the vector space V of all functions from R into R. (Furthermore,
P is a subspace of U , and consequently, a subspace of W as well.)
Example 31.
Let Pn be the set of all polynomials in the variable x of degree ≤ n. Then
under usual addition and scalar multiplication of polynomials, the set Pn is a
vector space. (Verify!). Let V = {p(x) ∈ Pn | p(1) = 0} be the set of all
polynomials in the variable x of degree ≤ n and vanishes at 1. Then V is a
subspace of Pn , since if p1 (x), p2 (x) ∈ V , then (p1 + p2 )(1) = p1 (1) + p2 (1) =
0 + 0 = 0, and (αp1 )(1) = αP1 (1) = α0 = 0, for all scalar α. But, the set
W = {p(x) ∈ Pn | p(x) is of degree n} is not a vector space as the zero polynomial
(the additive identity) ∈
/ W , and hence not a subspace of Pn .
Problem 55.
Let V be a vector space. If U and W be subspaces of V , then show that
U ∩ W is a subspace of V . Is U ∪ W a subspace of V ?
Solution.
Clearly, U ∩ W is a subset of V . Let u, v ∈ U ∩ W . Then u, v ∈ U and
u, v ∈ W . Since U and W are subspaces of V , u + v ∈ U , u + v ∈ W , αu ∈ U
and αu ∈ W , for any scalar α. Therefore u + v ∈ U ∩ W and αu ∈ U ∩ W .
Hence U ∩ W is a subspace of V . Note that the union of subspaces of a vector
space V need not be a subspace of V . For example, consider the vector space
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V = {(x, y, z)|x, y, z ∈ R}. Then W1 = {(0, 0, z)|z ∈ R} and W2 = {(0, y, 0)|y ∈
R} are subspaces of V . Now (0, 0, 3) and (0, 5, 0) are two elements of W1 ∪ W2 .
But their sum, (0, 5, 3) ∈
/ W1 ∪ W2 . Thus W1 ∪ W2 is not a subspace of V .
Problem 56.
Let V = {(x, y, z)|x, y, z ∈ R} and let W be subset of V consisting of all
(x, y, z) ∈ V such that x = 1 + y. Is W a subspace of V ?
Solution.
Clearly, V is a vector space with identity element (0, 0, 0). Since (0, 0, 0) ∈
/ W,
it can not be a vector space with respect to the operations in V , and hence not
a subspace of V .
Exercises.
1. If W1 and W2 are subspaces of a vector space V , then show that their union
is a subspace of V if and only if one of the spaces Wi is contained in the
other.
2. Let V be a vector space. Show that
(a) α0 = 0, for all scalar α.
(b) 0v = 0, for all v in V .
(c) (−1)v = −v, for all v in V .
(d) If α is a non zero scalar and v is a vector such that αv = 0, then v = 0
(e) If u, v are in V , and u + v = 0, then u = −v
3. Show that a non empty subset W of V is a subspace of V if and only if
for each pair of vectors u, v in W , and each scalar α, the vector αu + v is
again in W .
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4. Let V = {(x, y, z) | x, y, z ∈ R} and let W be the subset of V consisting of
all (x, y, z) ∈ V such that x + y + z = 0. Show that W is a subspace of V .
5. If W1 , W2 are subspaces of a vector space V , then show that their sum,
W = W1 + W2 = {w1 + w2 | w1 ∈ W1 , w2 ∈ W2 } is a subspace of V.
6. Let S = {p(x) ∈ P5 | p(1) = 0 and p(3) = 0}, where P5 is the space of all
polynomials in the variable x of degree ≤ 5. Show that S is a subspace of
P5 .
7. Let S = {p(x) ∈ Pn | p(x) = xp0 (x)}, where Pn is the space of all polynomials in the variable x of degree ≤ n and p0 (x) is the derivative of p(x). Is
S a subspace of Pn ?.
8. Let V be the set of all m × n matrices with real entries . For A = [aij ] and
B = [bij ] in V , and for a scalar α, define : A + B = [aij + bij ] and αA =
[αaij ]. Show that, under this rules of addition and scalar multiplication V
is a vector space.
6.2
Linear Dependence and Independence
Definition 6.2.1.
Let V be any vector space. A vector u ∈ V is said to be a linear combination of the vectors u1 , u2 , ..., un in V if there exist scalars α1 , α2 , ..., αn such
that u = α1 u1 + α2 u2 + ... + αn un .
For example, consider R3 , the space of all 3- tuples of real numbers. Note
that (3, 1, 5) = 3(1, 0, 0) + 1(0, 1, 0) + 5(0, 0, 1). Thus, we can say the vector
(3, 1, 5) in R3 is a linear combination of (1, 0, 0), (0, 1, 0) and (0, 0, 1).
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Remark.
If all scalars α1 , α2 , ..., αn are zeros, then the linear combination is called
trivial linear combination.
If at least one of αi0 s is non zero, then it is called a non-trivial linear
combination.
Definition 6.2.2.
Let V be a vector space , and let S = {u1 , u2 , ..., un } be a subset of V.
Then S is said to be linearly dependent(in short, LD) if there exist scalars
α1 , α2 , ..., αn , not all of which are zero, such that α1 u1 + α2 u2 + ... + αn un = 0.
If α1 = α2 = ... = αn = 0 is the only solution for the equation α1 u1 + α2 u2 +
... + αn un = 0, then S = {u1 , u2 , ..., un } is linearly independent (in short, LI).
Remark.
Any set which contains the zero vector is always linearly dependent (Why?).
Example 32.
Consider the vector space R3 . Then the set S = {v1 = (3, 0, −3), v2 =
(−1, 1, 2), v3 = (4, 2, −2), v4 = (2, 1, 1)} ⊂ R3 is linearly dependent, since 2v1 +
2v2 − v3 + 0.v4 = 0. The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) are
linearly independent, since α1 e1 + α2 e2 + α3 e3 = 0 =⇒ α1 = 0, α2 = 0, and
α3 = 0.
Remark.
An infinite subset S of a vector space V is said to be linearly dependent (resp.
independent) if every finite subset of S is linearly dependent (resp. independent).
Let S = {u1 , u2 , ..., un } be a subset of a vector space V. Now we will show
that the set of all linear combinations of vectors in S is a subspace of V. Let W
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denotes the set of all linear combinations of vectors u1 , u2 , ..., un . Then W ⊂ V .
Clearly, 0 = 0u1 + 0u2 + ... + 0un ∈ W =⇒ W is nonempty. Let u, v ∈ W .
Then u and v are linear combinations of u1 , u2 , ..., un . So, we can write u =
α1 u1 + α2 u2 + ... + αn un , v = β1 u1 + β2 u2 + ... + βn un where αi , βj are scalars.
We have u + v = (α1 u1 + α2 u2 + ... + αn un ) + (β1 u1 + β2 u2 + ... + βn un ) =
(α1 + β1 )u1 + (α2 + β2 )u2 + ... + (αn + βn )un ∈ W . For any scalar α, αu =
α(α1 u1 + α2 u2 + ... + αn un ) = (αα1 )u1 + (αα2 )u2 + ... + (ααn )un ∈ W . Hence by
theorem 6.1.3, W is a subspace of V . This subspace W is called the span of S,
and is denoted by [S].
Theorem 6.2.3.
(a) If a set S = {u1 , u2 , ..., un } of a vector space V is linearly independent, then
every subset of S is also linearly independent.
(b) If a set S = {u1 , u2 , ..., un } of a vector space V is linearly dependent, then
every superset of S is also linearly dependent.
Proof.
(a) Let S1 = {u1 , u2 , ..., uk } ⊂ S, where k ≤ n. Suppose that α1 u1 + α2 u2 + ... +
αk uk = 0. Then, we have α1 u1 + α2 u2 + ... + αk uk + 0uk+1 + ... + 0un = 0. Since
S is linearly independent, we get α1 = α2 = ... = αk = 0. Thus, S1 is linearly
independent.
(b) Given that S = {u1 , u2 , ..., un } is linearly dependent. Therefore, there are
scalars α1 , α2 , ..., αn , not all of which are zero, such that α1 u1 +α2 u2 +...+αn un =
0. Now let S1 = {u1 , u2 , ..., un , un+1 , ..., um } be any superset of S. Then we have
α1 u1 + α2 u2 + ... + αn un + 0un+1 + ... + 0um = 0, with not all α1 , α2 , ..., αn are
zero, showing that S1 is linearly dependent.
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Theorem 6.2.4.
If S = {u1 , u2 , ..., uk } is an ordered subset of nonzero elements of a vector space V , then S is linearly dependent if and only if one of the vectors of
{u1 , u2 , ..., uk } belongs to the span of the remaining vectors in S.
Proof.
Without loss of generality, assume that uk is a linear combination of
{u1 , u2 , ..., uk−1 }. Then, there exists scalars α1 , α2 , ..., αk−1 such that uk = α1 u1 +
α2 u2 + ... + αk−1 uk−1 . This implies, α1 u1 + α2 u2 + ... + αk−1 uk−1 − uk = 0. This
shows that S is linearly dependent.
Conversely, assume that S = {u1 , u2 , ..., uk } is linearly dependent . Then
there are scalars α1 , α2 , ..., αk , not all of which are zero, such that α1 u1 + α2 u2 +
... + αk uk = 0. Let αk 6= 0. Then, αk uk = −α1 u1 − α2 u2 − ... − αk−1 uk−1 , which
α2
αk−1
α1
uk−1 . Thus, uk is a linear combination
implies uk = − u1 − u2 − ... −
αk
αk
αk
of {u1 , u2 , ..., uk−1 }. Therefore, uk ∈ [u1 , u2 , ..., uk−1 ].
Problem 57.
Given that {u, v, w} is linearly independent, check whether {u−v, v−w, w−u}
is linearly independent or not.
Solution.
Let α(u − v) + β(v − w) + γ(w − u) = 0. This implies, (α − γ)u + (β − α)v +
(γ − β)w = 0. Since, {u, v, w} is linearly independent, we get α − γ = β − α =
γ − β = 0 =⇒ α = β = γ, which may be other than zero. Therefore the set
{u − v, v − w, w − u} is linearly dependent.
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Exercises.
1. Find the value of m such that (m, 7, −4) is a linear combination of the
vectors (−2, 2, 1) and (2, 1, −2).
2. Determine whether the set S = {1 + x, x + x2 , x2 + 1} ⊂ P2 , the space of
all polynomials of degree ≤ 2 is linearly dependent or independent.
3. Find three vectors in R3 which are linearly dependent and are such that
any two of them are linearly independent .
4. Show that the vectors v1 = (1, 0, −1), v2 = (1, 2, 1), and v3 = (0, 2, 2) in R3
are linearly dependent.
5. Prove that if two vectors in a vector space V are linearly dependent , one
of them is a scalar multiple of the other.
6. Are the vectors u1 = (1, 1, 2, 4), u2 = (2, −1, −5, 2), u3 = (1, −1, −4, 0) and
u4 = (2, 1, 1, 6) linearly dependent in R4 ?
7. Let V be a vector space and suppose α, β, γ are linearly independent vectors
in V . Prove that (α + β), (β + γ), (γ + α) are linearly independent.
6.3
Basis and Dimension
Definition 6.3.1.
Let V be a vector space. A subset B of V is called a basis of V , if (a) B is
linearly independent, and (b) [B] = V.
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Remark.
A vector space V may have different bases, but it can be shown that if one
basis of V has n elements , then any other basis of V also has n elements.
Definition 6.3.2.
Let B be a basis of a vector space V . If the number of vectors in B is n,
then the vector space V is called n-dimensional and we write dim(V ) = n. If
V consists of zero vector alone, then V does not have a basis, and we shall say
that V has dimension 0. A vector space V which has a basis consisting of a finite
number of elements, or the zero vector space, is called finite dimensional.
Other vector spaces are called infinite dimensional.
It can be proved that if V is a vector space, and if {v1 , v2 , ..., vn } is a basis of
V , then any elements w1 , w2 , ..., wm , with m > n of V , are linearly dependent.
In particular, if V is n-dimensional, any set of n + 1 vectors in V is linearly
dependent.
Example 33.
Consider the vector space Rn and consider the vectors e1 = (1, 0, ..., 0), ..., en =
(0, 0, ..., 1). Then, e1 , ..., en are linearly independent vectors in Rn . For, note
that α1 e1 + ... + αn en = 0 if and only if αi = 0 , for every i = 1, ..., n. Also, any
(x1 , x2 , ..., xn ) ∈ Rn can be written as: (x1, x2, ..., xn ) = x1 e1 + x2 e2 + ... + xn en .
Thus e1 , ..., en are linearly independent vectors in Rn and also [e1 , ..., en ] = Rn .
Thus, e1 , ..., en is a basis for Rn and hence Rn is an n-dimensional vector space.
Definition 6.3.3.
If U, W are subspaces of a vector space V , then the set of all sums, u + w of
vectors, where u ∈ U and w ∈ W is called the sum of the subspaces U and W
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and is denoted by U + W . Clearly, U + W is a subspace of V which contain each
of the subspaces U and W . Also, it can be proved that [U ∪ W ] = U + W .
Now, we state a theorem about the dimension of a sum of two subspaces of
a vector space.
Theorem 6.3.4.
If U and W are subspaces of an N -dimensional vector space V , then
dim(U + W ) = dim(U ) + dim(W ) − dim(U ∩ W ).
Proof. Let dim(U ) = m, dim(W ) = n, where m, n ≤ N . Since U ∩ W is a
subspace of both U and W , if dim(U ∩ W ) = r, then r ≤ both m and n.
Let B = {u1 , u2 , ..., ur } be a basis for U ∩ W .
Extend B to a basis for U , by adjoining (m − r) linearly independent vectors
v1 , v2 , ..., vm−r to B, so that B1 = {u1 , u2 , ..., ur , v1 , v2 , ..., vm−r } is a basis for U .
(See 8th question in the following exercise.)
Similarly, extend B to a basis for V , by adjoining (n − r) linearly independent
vectors w1 , w2 , ..., wn−r to B, so that B2 = {u1 , u2 , ..., ur , w1 , w2 , ..., wn−r } is a
basis for W .
Now, consider U + W = {u + w | u ∈ U, w ∈ W }.
Clearly, the vectors {u1 , u2 , ..., ur , v1 , v2 , ..., vm−r , w1 , w2 , ..., wn−r } spans the subspace U + W .
We claim that
{u1 , u2 , ..., ur , v1 , v2 , ..., vm−r , w1 , w2 , ..., wn−r }
is linearly independent.
Assume that α1 u1 + α2 u2 + ... + αr ur + β1 v1 + β2 v2 + ... + βm−r vm−r + γ1 w1 +
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γ2 w2 + ... + γn−r wn−r = 0
=⇒ α1 u1 + α2 u2 + ... + αr ur + β1 v1 + β2 v2 + ... + βm−r vm−r = −γ1 w1 − γ2 w2 −
... − γn−r wn−r
.............. (1)
=⇒ −γ1 w1 − γ2 w2 − ... − γn−r wn−r ∈ U .
Also, −γ1 w1 − γ2 w2 − ... − γn−r wn−r ∈ W.
Thus, −γ1 w1 − γ2 w2 − ... − γn−r wn−r ∈ U ∩ W , so that we can write
−γ1 w1 −γ2 w2 −...−γn−r wn−r = δ1 u1 +δ2 u2 +...+δr ur for some scalars δ1 , δ2 , ..., δr .
Therefore, γ1 w1 + γ2 w2 + ... + γn−r wn−r + δ1 u1 + δ2 u2 + ... + δr ur = 0.
Since, {u1 , u2 , ..., ur , w1 , w2 , ..., wn−r } is linearly independent, we see that γi = 0
and δj = 0, ∀i and ∀j.
Therefore, equation (1) becomes, α1 u1 + α2 u2 + ... + αr ur + β1 v1 + β2 v2 + ... +
βm−r vm−r = 0.
Since, {u1 , u2 , ..., ur , v1 , v2 , ..., vm−r } is linearly independent, we get αi = 0 and
βj = 0, ∀i, ∀j..
This shows that {u1 , u2 , ..., ur , v1 , v2 , ..., vm−r , w1 , w2 , ..., wn−r } is linearly independent and so it is a basis for U + W .
Hence dim(U + W ) = r + (m − r) + (n − r) = m + n − r = dim(U ) + dim(W ) −
dim (U ∩ W ).
Definition 6.3.5.
If V is a finite dimensional vector space, an ordered basis for V is a finite
sequence of vectors which is linearly independent and spans V . Suppose V is
n- dimensional vector space and B = {u1 , ..., un } is an ordered basis for V ,
then given v ∈ V , there exists a unique scalars (α1 , α2 , ..., αn ) such that v =
α1 u1 + α2 u2 + ... + αn un . Then (α1 , α2 , ..., αn ) is called the coordinate vector
of v relative to the ordered basis B.
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Problem 58.
Find the coordinates of (3, 4, 5) ∈ R3 relative to the ordered basis B =
{(1, 0, 1), (1, 1, 0), (0, 1, 1)} of R3 .
Solution.
Let (3, 4, 5) = α(1, 0, 1) + β(1, 1, 0) + γ(0, 1, 1). This implies, (3, 4, 5) =
(α + β, β + γ, α + γ). Equating both sides, we get α + β = 3, β + γ = 4, and
α + γ = 5. On solving, we get α = 2, β = 1, and γ = 3. Thus, the coordinates
of the vector (3, 4, 5) with respect to the given basis B is (2, 1, 3).
Exercises.
1. Show that S = {1, x, x2 } is a basis for P2 , the space of all polynomials of
degree ≤ 2.
2. Show that S = {(0, 1, 1), (1, 0, 1), (1, 1, 0)} is a basis for R3 .
3. Let V be a vector space of dimension n, and let v1 , v2 , .., vn be linearly
independent elements of V . Then show that v1 , v2 , ..., vn constitutes a basis
for V .
4. Consider the subspace U = {(x, y, z)|x − y + z = 0} of R3 . Find a basis of
U . What is dim U ?
5. Let V be an n-dimensional vector space. Then show that
(a) Any subset of V which contains more than n vectors is linearly dependent;
(b) No subset of V which contains less than n vectors can span V .
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6. Let S be a linearly independent subset of a vector space V . Suppose β
is a vector in V which is not in [S]. Then show that the set obtained by
adjoining β to S is linearly independent.
7. If W is a subspace of a finite dimensional vector space V , then show that
every linearly independent subset of W is finite and is a part of a basis for
W.
8. Let V be a vector space of dimension n. Let r be a positive integer with
r < n, and let v1 , v2 , ..., vr be linearly independent elements of V . Then
show that there exists elements vr+1 , vr+2 , ..., vn−r such that {v1 , v2 , ..., vn }
is a basis of V.
9. Find the coordinates of (−1, 2, 3) relative to the ordered basis
B = {(1, 1, 1), (2, 0, 1), (2, 3, 5)} of R3 .
10. Find the coordinates of x2 + 2x − 1 relative to the ordered basis
B = {x + 1, x2 + x − 1, x2 − x + 1} of P2 .
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Text Books (As per Syllabus)
1. John B. Fraleigh: A First Course in Abstract Algebra, 7th Ed., Pearson
Education Inc.
2. D. Prasad: Linear Algebra, Narosa Pub. House
Further Reading
1. Joseph A. Gallian: Contemporary Abstract Algebra, Narosa Pub. House.
2. I. N. Herstein: Topics in Algebra, 2nd Edition, John Wiley and Sons.
3. Kenneth Hoffman, and Ray Kunze: Linear Algebra ,2nd Edition, Prentice
Hall of India Private Ltd.
4. Artin: Algebra, Prentice Hall of India Private Ltd.
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